From 7b1e0f329e7e31f8051ce9fd05a4cf6642462259 Mon Sep 17 00:00:00 2001 From: xingyou wu <3050128610@qq.com> Date: Fri, 14 Mar 2025 22:39:32 +0800 Subject: [PATCH] Initial commit --- .DS_Store | Bin 0 -> 6148 bytes 14lanqiao/test1.cpp | 28 +++++++++ 14lanqiao/test2.cpp | 13 ++++ 14lanqiao/test3-1.cpp | 49 +++++++++++++++ 14lanqiao/test3-2.cpp | 26 ++++++++ 14lanqiao/test4.cpp | 50 ++++++++++++++++ 14lanqiao/test5-1.cpp | 33 +++++++++++ 14lanqiao/test5-2.cpp | 25 ++++++++ 15lanqiao/test1-1.cpp | 15 +++++ 15lanqiao/test1-2.cpp | 17 ++++++ 15lanqiao/test2.cpp | 19 ++++++ 15lanqiao/test3.cpp | 30 ++++++++++ 15lanqiao/test4.cpp | 34 +++++++++++ 15lanqiao/test5.cpp | 45 ++++++++++++++ 15lanqiao/test6.cpp | 57 ++++++++++++++++++ 15lanqiao/test7.cpp | 26 ++++++++ 15lanqiao/test8.cpp | 44 ++++++++++++++ test100.cpp | 9 +++ test99-1.cpp | 134 ++++++++++++++++++++++++++++++++++++++++++ test99-2.cpp | 44 ++++++++++++++ test99-3-1.cpp | 34 +++++++++++ test99-3-2.cpp | 31 ++++++++++ test99-3-3.cpp | 44 ++++++++++++++ test99-3-4 | Bin 0 -> 35144 bytes test99-3-4.cpp | 33 +++++++++++ test99-3-5 | Bin 0 -> 51768 bytes test99-3-5.cpp | 59 +++++++++++++++++++ test99-4-1 | Bin 0 -> 49768 bytes test99-4-1-1.cpp | 44 ++++++++++++++ test99-4-1-2.cpp | 45 ++++++++++++++ test99-4-2 | Bin 0 -> 34712 bytes test99-4-2.cpp | 37 ++++++++++++ test99-4-3 | Bin 0 -> 34840 bytes test99-4-3.cpp | 39 ++++++++++++ test99-4-4 | Bin 0 -> 34776 bytes test99-4-4.cpp | 23 ++++++++ test99-4-5 | Bin 0 -> 51688 bytes test99-4-5.cpp | 66 +++++++++++++++++++++ 38 files changed, 1153 insertions(+) create mode 100644 .DS_Store create mode 100644 14lanqiao/test1.cpp create mode 100644 14lanqiao/test2.cpp create mode 100644 14lanqiao/test3-1.cpp create mode 100644 14lanqiao/test3-2.cpp create mode 100644 14lanqiao/test4.cpp create mode 100644 14lanqiao/test5-1.cpp create mode 100644 14lanqiao/test5-2.cpp create mode 100644 15lanqiao/test1-1.cpp create mode 100644 15lanqiao/test1-2.cpp create mode 100644 15lanqiao/test2.cpp create mode 100644 15lanqiao/test3.cpp create mode 100644 15lanqiao/test4.cpp create mode 100644 15lanqiao/test5.cpp create mode 100644 15lanqiao/test6.cpp create mode 100644 15lanqiao/test7.cpp create mode 100644 15lanqiao/test8.cpp create mode 100644 test100.cpp create mode 100644 test99-1.cpp create mode 100644 test99-2.cpp create mode 100644 test99-3-1.cpp create mode 100644 test99-3-2.cpp create mode 100644 test99-3-3.cpp create mode 100755 test99-3-4 create mode 100644 test99-3-4.cpp create mode 100755 test99-3-5 create mode 100644 test99-3-5.cpp create mode 100755 test99-4-1 create mode 100644 test99-4-1-1.cpp create mode 100644 test99-4-1-2.cpp create mode 100755 test99-4-2 create mode 100644 test99-4-2.cpp create mode 100755 test99-4-3 create mode 100644 test99-4-3.cpp create mode 100755 test99-4-4 create mode 100644 test99-4-4.cpp create mode 100755 test99-4-5 create mode 100644 test99-4-5.cpp diff --git a/.DS_Store b/.DS_Store new file mode 100644 index 0000000000000000000000000000000000000000..020e7d6b589c6bbd36bcbc0919aeab787f86deb6 GIT binary patch literal 6148 zcmeHKu};H44E2=`saPt+z`$@TV5jSTSJjVPsKDsgX(`%9;@P%S0p{pWtKD&%=lh61a}-o@j~(d*mUCeQgVJ(t(b zlVbv$-$^^a`TeXcLMS=Ep5=1U+~n}R->aGa>f?0PNdMb$RrpH(XU}%}Pn@5LF<=ZB z16#)cYBo!+DHf!L2g Mq`?Mb;7=L&1nOH^i2wiq literal 0 HcmV?d00001 diff --git a/14lanqiao/test1.cpp b/14lanqiao/test1.cpp new file mode 100644 index 0000000..6c70da6 --- /dev/null +++ b/14lanqiao/test1.cpp @@ -0,0 +1,28 @@ +#include +using namespace std; +const int N = 1e2 + 10; +int a[N] = { 0,5,6,8,6,9,1,6,1,2,4,9,1,9,8,2,3,6,4,7,7,5,9,5,0,3,8,7,5,8,1,5,8,6,1,8,3,0,3,7,9,2, +7,0,5,8,8,5,7,0,9,9,1,9,4,4,6,8,6,3,3,8,5,1,6,3,4,6,7,0,7,8,2,7,6,8,9,5,6,5,6,1,4,0,1, +0,0,9,4,8,0,9,1,2,8,5,0,2,5,3,3 }; // 示例数组 + +int day[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; +int main(){ + int ans = 0; + for(int i = 1; i <=12; i++){ + for(int j = 1; j <= day[i]; j++){ + int date[] = {0, 2, 0, 2, 3, i/10, i%10, j/10, j%10 }; + int id = 1; + for(int i = 1; i<=100; i++){ + if(a[i] == date[id]){ + id++; + if(id>8){ + ans++; + break; + } + } + } + } + } + cout << ans << endl; + return 0; +} \ No newline at end of file diff --git a/14lanqiao/test2.cpp b/14lanqiao/test2.cpp new file mode 100644 index 0000000..3b943d9 --- /dev/null +++ b/14lanqiao/test2.cpp @@ -0,0 +1,13 @@ +#include +using namespace std; +int main(){ + double n = 23333333; + double res = 11625907.5798; + for(int a = 0; a <= n; a++){ // a为0出现的次数 + int b = n - a; // b为1出现的次数 + if(a>=b) continue; + double ans = (-1)*a*(a/n)*log2(a/n) + (-1)*b*(b/n)*log2(b/n); + if(fabs(ans - res) < 1e-4) cout << a << endl; + } + return 0; +} \ No newline at end of file diff --git a/14lanqiao/test3-1.cpp b/14lanqiao/test3-1.cpp new file mode 100644 index 0000000..fc5689e --- /dev/null +++ b/14lanqiao/test3-1.cpp @@ -0,0 +1,49 @@ +#include +using namespace std; +#define int long long +const int N = 1e4 + 10; +int a[N], b[N]; +//枚举法 +signed main(){ + int n; cin >> n; + int mxv = LLONG_MIN, miv = LLONG_MAX; + for(int i = 1; i <= n; i++){ + cin >> a[i] >> b[i]; + if(b[i] != 1) mxv = max(mxv, a[i]/(b[i] - 1)); + miv = min(miv, a[i]/(b[i] + 1)); + } + + for(int v = miv; v <= mxv; v++){ + bool flag = true; + for(int i = 1; i <= n; i++){ + if(a[i] / v != b[i]){ + flag = false; + break; + } + } + if(flag){ + cout << v << " "; + break; + } + } + for(int v = mxv; v >= miv; v--){ + bool flag = true; + for(int i = 1; i <= n; i++){ + if(a[i] / v != b[i]){ + flag = false; + break; + } + } + if(flag){ + cout << v << endl; + break; + } + } + return 0; +} +/* test samples +3 +75 3 +53 2 +59 2 +*/ \ No newline at end of file diff --git a/14lanqiao/test3-2.cpp b/14lanqiao/test3-2.cpp new file mode 100644 index 0000000..086989d --- /dev/null +++ b/14lanqiao/test3-2.cpp @@ -0,0 +1,26 @@ +#include +using namespace std; +int main(){ + int n; cin >> n; + int mxv = INT_MIN; int miv = INT_MAX; + for(int i = 1; i <= n; i++){ + int a, b; cin >> a >> b; + miv = min(miv, a/b); + mxv = max(mxv, a/(b+1)+1); + } + cout << mxv << " " << miv << endl; + return 0; +} +/* +数学优化 +通过floor(A/V) = B，可以推出 A/V =［B,B+1） +进而得到B <= A/V < B+1，三个表达式同时被A除 +得到关于V的不等式即 A/B >= V > A/(B+1) +*/ + +/* test samples +3 +75 3 +53 2 +59 2 +*/ \ No newline at end of file diff --git a/14lanqiao/test4.cpp b/14lanqiao/test4.cpp new file mode 100644 index 0000000..f028207 --- /dev/null +++ b/14lanqiao/test4.cpp @@ -0,0 +1,50 @@ +#include +using namespace std; +const int N = 1e2 + 10; +struct node{ + int t, d, l; // 分别代表可以降落的时刻，盘旋时间，降落花费时间 +}a[N]; +bool vis[N]; +int t, n; +bool flag = false; +// lasttime: 之前完成降落的所有飞机中的最后一架飞机降落的时间 +void dfs(int dep, int lasttime){ + if(dep == n + 1){ + flag = true; + return; + } + for(int i = 1; i <= n; i++){ + if(!vis[i] && a[i].t + a[i].d >= lasttime){ + vis[i] = 1; + dfs(dep + 1, max(lasttime, a[i].t) + a[i].l); + // 遍历同级的另一个飞机的下一个深度的飞机前，需要将该飞机设置为未安排状态 + vis[i] = 0; + } + } +} +int main(){ + cin >> t; + while(t--){ + cin >> n; + for(int i = 1; i<=n; i++){ + cin >> a[i].t >> a[i].d >> a[i].l; + } + flag = false; + memset(vis, 0, sizeof(vis)); + dfs(1, 0); + if(flag) cout << "YES" << endl; + else cout << "NO" << endl; + } + return 0; +} +/* test samples +2 +3 +0 100 10 +10 10 10 +0 2 20 +3 +0 10 20 +10 10 20 +20 10 20 +*/ \ No newline at end of file diff --git a/14lanqiao/test5-1.cpp b/14lanqiao/test5-1.cpp new file mode 100644 index 0000000..08aa93b --- /dev/null +++ b/14lanqiao/test5-1.cpp @@ -0,0 +1,33 @@ +#include +using namespace std; +const int N = 1e5 + 10; +int ft[N], bk[N], dp[N]; +/* +ft[i]第i个元素的首数字、bk［i］第i个元素尾数字 +类LIS问题的DP解法-时间复杂度0(n^2) +状态 dp[i] 以第主个元素作为结尾的最长接龙数列的长度 +状态转移方程 dp[i] = max(dp[i], dp[j]+1); +*/ +int main(){ + int n; cin >> n; + string s; + for(int i = 1; i<=n; i++){ + cin >> s; + ft[i] = s.front() - '0', bk[i] = s.back() - '0', dp[i] = 1; + } + int mx = 1; + for(int i = 2; i <= n; i++){ + for(int j = 1; j < i; j++){ + if(bk[j] == ft[i]){ + dp[i] = max(dp[i], dp[j] + 1); + } + mx = max(mx, dp[i]); + } + } + cout << n - mx << endl; + return 0; +} +/* test samples 易超时 +5 +11 121 22 12 2023 +*/ \ No newline at end of file diff --git a/14lanqiao/test5-2.cpp b/14lanqiao/test5-2.cpp new file mode 100644 index 0000000..617338d --- /dev/null +++ b/14lanqiao/test5-2.cpp @@ -0,0 +1,25 @@ +#include +using namespace std; +int dp[10]; +/* +优化dp方案 +状态 dp[i] 以数字i结尾的最长接龙序列的长度 +状态转移方程 dp[bk]=max(dp[ft]+1, dp[bk]) +*/ +int main(){ + int n; cin >> n; + string s; + int mx = 1; + for(int i = 0; i < n; i++){ + cin >> s; + int ft = s.front() - '0', bk = s[s.size() - 1] - '0'; + dp[bk] = max(dp[ft] + 1, dp[bk]); + mx = max(mx, dp[bk]); + } + cout << n - mx << endl; + return 0; +} +/* test samples +5 +11 121 22 12 2023 +*/ \ No newline at end of file diff --git a/15lanqiao/test1-1.cpp b/15lanqiao/test1-1.cpp new file mode 100644 index 0000000..258e8e1 --- /dev/null +++ b/15lanqiao/test1-1.cpp @@ -0,0 +1,15 @@ +#include +using namespace std; + +int main() { + int tot = (50*49) / 2, cnt = (7*6) / 2; + cout << tot - cnt << endl; + return 0; +} + +/* +总共有50人参与会议，如果按照常规情况，即每个人与除了自己以外的所有人各握手一次 +握手总数tot=49+48+...+1=(50*49)/2 +现在有7个人没有互相握手，这7这个人本来应该握手的次数 cnt=6+5+...+1=(7*6)/2 +实际握手次数=总的握手次数tot-7人的情况cnt (tot - cnt) +*/ \ No newline at end of file diff --git a/15lanqiao/test1-2.cpp b/15lanqiao/test1-2.cpp new file mode 100644 index 0000000..8982a24 --- /dev/null +++ b/15lanqiao/test1-2.cpp @@ -0,0 +1,17 @@ +#include +using namespace std; + +int main() { + int ans = 0; + for(int i = 1; i <=50; i++){ + for(int j = i + 1; j<=50; j++){ + if(!( (i >= 1 && i<= 7) && (j >= 1 && j <= 7) )) ans++; + // i和j不能同时满足都在[1, 7]范围内 + } + } + cout << ans << endl; + return 0; +} +/* 枚举法 + +*/ \ No newline at end of file diff --git a/15lanqiao/test2.cpp b/15lanqiao/test2.cpp new file mode 100644 index 0000000..4014a75 --- /dev/null +++ b/15lanqiao/test2.cpp @@ -0,0 +1,19 @@ +#include +using namespace std; + +int main() { + int x = 343720, y = 233333, dx = 15, dy = 17; + int p = y*dx, q = x*dy; + int g = gcd(p, q); + p /= g, q /= g; + int t = 2*p*x / dx; + double ans = t * (sqrt(15*15 + 17*17)); + printf("%.2lf\n", ans); + return 0; +} +/* +思路:针对前进的方向进行分解，分为x方向的运动和y方向的运动假设x方向走了p个来回，y反向走了q个来回， +经过了时间t，小球第一次回到原点则t*dx=2px,t*dy=2qy，结合1式/2式，得p/q=y/x*dx/dy +利用gcd对分式p,q进行约分，进而得到约分后的p,q +则时间 t = 2px/dx, 总路程 = t * sqrt(15*15 + 17*17) +*/ \ No newline at end of file diff --git a/15lanqiao/test3.cpp b/15lanqiao/test3.cpp new file mode 100644 index 0000000..fb13dde --- /dev/null +++ b/15lanqiao/test3.cpp @@ -0,0 +1,30 @@ +#include +using namespace std; +bool check(int n){ + int b = 1; //数位 + while(n){ + if(b % 2){ // 奇数位 + if(!(n % 10 % 2)) return false; + }else{ // 偶数位 + if(n % 10 % 2) return false; + } + n /= 10; + b++; + } + return true; +} + +int main() { + int n; cin >>n; + int ans = 0; + for(int i = 1; i <= n; i++){ + if(i % 10 % 2 == 0) continue; + if(check(i)) ans++; + } + cout << ans << endl; + return 0; +} +/* +一个整数如果按从低位到高位的顺序，奇数位为奇数，偶数位为偶数，则该数为好数 +24 -> 7, 2024 -> 150 +*/ \ No newline at end of file diff --git a/15lanqiao/test4.cpp b/15lanqiao/test4.cpp new file mode 100644 index 0000000..3c0d00c --- /dev/null +++ b/15lanqiao/test4.cpp @@ -0,0 +1,34 @@ +#include +using namespace std; +const int N = 2e3 + 10; +string s; int a[N]; +int main() { + int n; cin >> n >> s; + reverse(s.begin(), s.end()); + int pos = s.find("."); // 小数点位置 + s.erase(pos, 1); // 删除小数点, 方便后续计算 + int len = s.size(); + for(int i = 0; i < len; i++) a[i+1] = s[i] - '0'; +// a[1] = 4，a[2] = 1，a[3] = 3 + + //高精度*低精度模版 + for(int i = 1; i<=n; i++){ + // 顺序扫描每一位, 均乘2 + for(int j = 1; j <= len; j++) a[j] *= 2; + for(int j = 1; j <= len; j++){ + if(a[j] > 10){ + a[j+1]++; + a[j] %= 10; + if(j == len) len++; + } + } + } +// a[1] = 6，a[2] = 5，a[3] = 2，a[4] = 1 a[pos] = 5 + if(a[pos] >= 5) a[pos+1]++; + for(int i = len; i >= pos + 1; i--) cout << a[i]; + cout << endl; + return 0; +} +/* +2 3.14 +*/ \ No newline at end of file diff --git a/15lanqiao/test5.cpp b/15lanqiao/test5.cpp new file mode 100644 index 0000000..01dd4fe --- /dev/null +++ b/15lanqiao/test5.cpp @@ -0,0 +1,45 @@ +#include +using namespace std; +const int N = 1e5 + 10; +int a[N]; +vector fac[N], s[N]; + +int main() { + for(int i = 1; i <= 1e5; i++){ + for(int j = i; j <= 1e5; j += i){ + fac[j].push_back(i); + } + } + int n; cin>>n; + for(int i = 1; i<=n; i++) cin >> a[i]; + // 保证字典序最小 + sort(a+1, a+1+n); + for(int i = 1; i <=n; i++){ + for(int j = 0; j < fac[a[i]].size(); j++){ + s[fac[a[i]][j]].push_back(a[i]); + } + } + for(int i = 1e5; i>=0; i--){ + if(s[i].size() >= 3){ + cout << s[i][0] << " " << s[i][1] << " " << s[i][2] << endl; + break; + } + } + return 0; +} +/* test samples +5 +1 2 3 4 9 +*/ + +/* +排序后的宝石为 [1, 2, 3, 4, 9]。 +因数数组 fac 和宝石数组 s 的关系如下： +s[1] 包含所有宝石 [1, 2, 3, 4, 9]。 +s[2] 包含 [2, 4]。 +s[3] 包含 [3, 9]。 +s[4] 包含 [4]。 +s[9] 包含 [9]。 +最终，s[1] 中有5个宝石，满足条件。 +输出前三个宝石：1 2 3。 +*/ \ No newline at end of file diff --git a/15lanqiao/test6.cpp b/15lanqiao/test6.cpp new file mode 100644 index 0000000..8efd5a1 --- /dev/null +++ b/15lanqiao/test6.cpp @@ -0,0 +1,57 @@ +#include +using namespace std; +const int N = 20; +int a[N][N]; // 迷宫数组 +int vis[N][N]; // 标记数组 +int n, k; + // 方向数组 0 1 2 3 4 5 6 7 +int dx[] = {-1, -1, 0, 1, 1, 1, 0, -1}; +int dy[] = { 0, 1, 1, 1, 0, -1, -1, -1}; +string res; // 结果字符串 + +void dfs(int x, int y, int pre, string s, int dep){ + if(x == n && y == n && dep == n*n){ + if(res.empty()) res = s; // 字典需最小的 + return; + } + for(int i = 0; i < 8; i++){ + int bx = x + dx[i], by = y + dy[i]; + if(bx < 1 || bx > n || by < 1 || by > n) continue; + if(vis[bx][by]) continue; + // 防止交叉搜索 + if(i == 1 && vis[x-1][y] && vis[x][y+1]) continue; + else if(i == 3 && vis[x+1][y] && vis[x][y+1]) continue; + else if(i == 5 && vis[x+1][y] && vis[x][y-1]) continue; + else if(i == 7 && vis[x-1][y] && vis[x][y-1]) continue; + // 保证路径数值为0 1 ... k-1 + if((a[bx][by] < k && a[bx][by] == pre + 1) || (pre + 1 == k && a[bx][by] == 0)){ + vis[bx][by] = 1; + dfs(bx, by, a[bx][by], s + to_string(i), dep + 1); + // 最优性剪枝 + if(!res.empty()) return; + vis[bx][by] = 0; //回溯 + } + } +} + +int main() { + cin >> n >> k; + for(int i = 1; i <= n; i++){ + for(int j = 1; j <= n; j++){ + cin >> a[i][j]; + } + } + string emp; + vis[1][1] = 1; + dfs(1, 1, 0, emp, 1); + if(res.empty()) cout << -1 << endl; + else cout << res << endl; + return 0; +} + +/* test samples +3 3 +0 2 0 +1 1 1 +2 0 2 +*/ \ No newline at end of file diff --git a/15lanqiao/test7.cpp b/15lanqiao/test7.cpp new file mode 100644 index 0000000..8454f54 --- /dev/null +++ b/15lanqiao/test7.cpp @@ -0,0 +1,26 @@ +#include +using namespace std; +const int N = 1e5 + 10; +int main(){ + priority_queue pq; // 大根堆 + int n, p ,q; cin >> n >> p >> q; + for(int i = 1; i <= n; i++){ + int x; cin >> x; + pq.push(x); + } + while(p || q){ + int cur = pq.top(); // 最大值 + pq.pop(); + if(p){ cur = sqrt(cur); p--; } + else if(q) { cur = cur / 2; q--; } + pq.push(cur); + } + int s = 0; // 总的消耗 + while(!pq.empty()) { s += pq.top(); pq.pop(); } + cout << s << endl; + return 0; +} +/* test samples +4 1 1 +4 5 6 49 +*/ \ No newline at end of file diff --git a/15lanqiao/test8.cpp b/15lanqiao/test8.cpp new file mode 100644 index 0000000..9d1d97b --- /dev/null +++ b/15lanqiao/test8.cpp @@ -0,0 +1,44 @@ +#include +using namespace std; +#define int long long +const int N = 1e3+10; +int a[N], s[N]; // s为前缀和数组 +multiset ms; + +signed main(){ + int n; cin>>n; + for(int i = 1; i<=n; i++){ + cin >> a[i]; + s[i] = s[i-1] + a[i]; + } + // 使用set数组去维护所有的区间和 + for(int i = 1; i<=n; i++){ + for(int j = 1; j<=n; j++){ + ms.insert(s[j] - s[i-1]); + } + } + int ans = LLONG_MAX; + + // 时间复杂度为 O(n^log2n) + for(int i = 1; i<=n; i++){ + for(int j = 1; j +using namespace std; + +int main(){ + int n; + n++; + cout << n << endl; + return 0; +} \ No newline at end of file diff --git a/test99-1.cpp b/test99-1.cpp new file mode 100644 index 0000000..46076da --- /dev/null +++ b/test99-1.cpp @@ -0,0 +1,134 @@ +#include +using namespace std; + +typedef struct LNode{ + int data; + struct LNode *next; +}LNode, *LinkList; + +bool InitList(LinkList &L){ + L = (LNode *)malloc(sizeof(LNode)); // 带头节点 + if(L == NULL) return false; + L->next = NULL; + return true; +} + +void InputList(LinkList &L){ + int x; cin>>x; + LinkList tail = L; + while(x != -99999){ // 输入 -99999 结束输入 + LNode *newNode = (LNode*)malloc(sizeof(LNode)); + newNode->data = x; newNode->next = NULL; + tail->next = newNode; + tail = newNode; + cin >> x; + } +} + +void PrintList(LinkList &L){ + LinkList p = L->next; + while(p != NULL){ + cout << p->data << ",\n"[p->next == NULL]; + p = p->next; + } +} + +LinkList p1, p2, p3, tmp; +void MergeList1(LinkList &L1, LinkList &L2, LinkList &L3){ + // 升序合并并去重 + p1 = L1->next; p2 = L2->next; + L3 = p3 = L1; + while(p1 && p2){ + if(p1->data < p2->data){ + p3->next = p1; p3 = p1; p1 = p1->next; + }else if(p2->data < p1->data){ + p3->next = p2; p3 = p2; p2 = p2->next; + }else{ + p3->next = p1; p3 = p1; p1 = p1->next; + tmp = p2->next; delete p2; p2 = tmp; + } + } + p3->next = p1?p1:p2; + delete L2; +} + +void MergeList2(LinkList &L1, LinkList &L2, LinkList &L3){ + // 降序合并 + p1 = L1->next; p2 = L2->next; + L3 = L1; + L3->next = NULL; + while(p1 || p2){ + if(!p1) { tmp = p2; p2 = p2->next; } + else if(!p2) { tmp = p1; p1 = p1->next; } + else if(p1->data <= p2->data) { tmp = p1; p1 = p1->next; } + else { tmp = p2; p2 = p2->next; } + + tmp->next = L3->next; L3->next = tmp; + } + delete L2; +} + +void MixList(LinkList &L1, LinkList &L2, LinkList &L3){ + // 两链表相交 + p1 = L1->next; p2 = L2->next; + L3 = p3 = L1; + while(p1 && p2){ + if(p1->data == p2->data){ + p3->next = p1; p3 = p1; p1 = p1->next; + tmp = p2; p2 = p2->next; delete tmp; + }else if(p1->data < p2->data){ + tmp = p1; p1 = p1->next; delete tmp; + }else{ + tmp = p2; p2 = p2->next; delete tmp; + } + } + while(p1) { tmp = p1; p1 = p1->next; delete tmp; } + while(p2) { tmp = p2; p2 = p2->next; delete tmp; } + + p3->next = NULL; + delete L2; +} + +void DiffList(LinkList &L1, LinkList &L2, int &n){ + // 两链表相差 + p1 = L1->next; p2 = L2->next; + p3 = L1; + while(p1 && p2){ + if(p1->data < p2->data){ + p3 = p1; p1 = p1->next; n++; + }else if(p1->data > p2->data){ + p2 = p2->next; + }else{ + p3->next = p1->next; + tmp = p1; p1 = p1->next; delete tmp; + } + } +} + +int main(){ + LinkList L1, L2, L3; InitList(L1); InitList(L2); InitList(L3); + InputList(L1); InputList(L2); // PrintList(L1); PrintList(L2); // 初始化并输入输出两个链表 + MergeList1(L1, L2, L3); + cout << "两链表升序合并并去重: "; PrintList(L3); + + LinkList L4, L5, L6; InitList(L4); InitList(L5); InitList(L6); + InputList(L4); InputList(L5); // PrintList(L4); PrintList(L5); // 初始化并输入输出两个链表 + MergeList2(L4, L5, L6); + cout << "两链表降序合并: "; PrintList(L6); + + LinkList L7, L8, L9; InitList(L7); InitList(L8); InitList(L9); + InputList(L7); InputList(L8); + MixList(L7, L8, L9); + cout << "两链表相交: "; PrintList(L9); + + LinkList L10, L11; InitList(L10); InitList(L11); + InputList(L10); InputList(L11); int n = 0; + DiffList(L10, L11, n); + cout << "两链表相差: "; PrintList(L10); + cout << "元素个数位: " << n << endl; + return 0; +} +/* test samples +1 3 5 7 9 13 15 -99999 +2 4 6 8 10 11 13 14 15 -99999 +*/ \ No newline at end of file diff --git a/test99-2.cpp b/test99-2.cpp new file mode 100644 index 0000000..ecc4855 --- /dev/null +++ b/test99-2.cpp @@ -0,0 +1,44 @@ +// 主元素寻找 +#include +#include + +using namespace std; + +int findMajorityElement(const vector& nums) { + int candidate = -1; + int count = 0; + + // 第一次遍历：找出候选主元素 + for (int num : nums) { + if (count == 0) { + candidate = num; + } + count += (num == candidate) ? 1 : -1; + } + + // 第二次遍历：验证候选主元素 + count = 0; + for (int num : nums) { + if (num == candidate) { + count++; + } + } + + // 如果候选元素出现次数超过数组长度的一半，则为所求主元素 + if (count > nums.size() / 2) { + return candidate; + } else { + return -1; + } +} + +int main() { + vector nums = {0, 5, 3, 5, 5, 7, 5, 5}; + int result = findMajorityElement(nums); + if (result != -1) { + cout << "主元素是: " << result << endl; + } else { + cout << "没有主元素" << endl; + } + return 0; +} \ No newline at end of file diff --git a/test99-3-1.cpp b/test99-3-1.cpp new file mode 100644 index 0000000..26cdea1 --- /dev/null +++ b/test99-3-1.cpp @@ -0,0 +1,34 @@ +#include +using namespace std; + +vector> results; + +void backtrack(vector& current, vector& used, int n){ + if(current.size() == n){ + results.push_back(current); + return; + } + for(int i = 1; i <= n; i++){ + if(!used[i]){ + current.push_back(i); + used[i] = true; + backtrack(current, used, n); + current.pop_back(); + used[i] = false; + } + } +} + +int main(){ + int n; cin >> n; + vector current; + vector used(n+1, false); + backtrack(current, used, n); + for(const auto& perm:results){ + for(int num:perm){ + cout << num << " "; + } + cout << endl; + } + return 0; +} \ No newline at end of file diff --git a/test99-3-2.cpp b/test99-3-2.cpp new file mode 100644 index 0000000..1424374 --- /dev/null +++ b/test99-3-2.cpp @@ -0,0 +1,31 @@ +#include +using namespace std; + +const int N = 1e2 + 10; // 定义最大数组大小 +int a[N]; // 存储排列结果 +bool vis[N]; // 标记数字是否被访问过 +int n; // 排列的大小 + +// 回溯函数，dep表示当前处理到第几个数字 +void dfs(int dep) { + if (dep == n + 1) { // 如果已经处理完所有数字 + for (int i = 1; i <= n; i++) { // 输出排列 + cout << a[i] << " \n"[i==n]; + } + return; + } + + for (int i = 1; i <= n; i++) { // 尝试将1到n的每个数字放入当前位置 + if (vis[i]) continue; // 如果该数字已经被使用，跳过 + vis[i] = true; // 标记该数字为已使用 + a[dep] = i; // 将数字放入当前位置 + dfs(dep + 1); // 递归处理下一个位置 + vis[i] = false; // 回溯，恢复状态 + } +} + +int main() { + cin >> n; // 输入排列的大小 + dfs(1); // 从第一个位置开始递归 + return 0; +} \ No newline at end of file diff --git a/test99-3-3.cpp b/test99-3-3.cpp new file mode 100644 index 0000000..8a50380 --- /dev/null +++ b/test99-3-3.cpp @@ -0,0 +1,44 @@ +// lanqiao1508 N皇后问题 +#include +using namespace std; +const int N = 15; +int n, ans; +int vis[N][N]; + +void dfs(int dep){ + if(dep == n + 1){ + ans++; + return; + } + for(int i = 1; i <= n; i++){ // 为什么下面只标记同一行，因为每次搜寻时只会在同一行(深度)寻找一个皇后 + if(vis[dep][i]) continue; // 所以只需要排除某行某个元素的同一列即可 + // 修改状态 + for(int _i = 1; _i <= n; _i++) vis[_i][i]++; // 标记同一列 + // 表壳 X 形模版 + for(int _i = dep, _j = i; _i >= 1 && _j >= 1; _i--, _j--) vis[_i][_j]++; // 标记左上对角线 + for(int _i = dep, _j = i; _i <= n && _j >= 1; _i++, _j--) vis[_i][_j]++; // 标记右上对角线 + for(int _i = dep, _j = i; _i >= 1 && _j <= n; _i--, _j++) vis[_i][_j]++; // 标记左下对角线 + for(int _i = dep, _j = i; _i <= n && _j <= n; _i++, _j++) vis[_i][_j]++; // 标记右下对角线 + + // 搜查下一深度 + dfs(dep + 1); + + // 恢复现场 + for(int _i = 1; _i <= n; _i++) vis[_i][i]--; + for(int _i = dep, _j = i; _i >= 1 && _j >= 1; _i--, _j--) vis[_i][_j]--; + for(int _i = dep, _j = i; _i <= n && _j >= 1; _i++, _j--) vis[_i][_j]--; + for(int _i = dep, _j = i; _i >= 1 && _j <= n; _i--, _j++) vis[_i][_j]--; + for(int _i = dep, _j = i; _i <= n && _j <= n; _i++, _j++) vis[_i][_j]--; + } +} + +int main(){ + ios::sync_with_stdio(0), cin.tie(0), cout.tie(0); + cin >> n; + dfs(1); + cout << ans << '\n'; + return 0; +} +/* test samples +5 +*/ \ No 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mp[i][j] == '.') continue; + scc++; // scc表示当前颜色编号 + dfs(i, j); + } + } + int ans = 0; + for(int i = 1; i <= n; i++){ + for(int j = 1; j <= n; j++){ + if(mp[i][j] == '.') continue; + bool tag = true; + for(int k = 0; k < 4; k++){ + int x = i + dx[k], y = j + dy[k]; + if(mp[x][y] == '.') tag = false; + } + if(tag){ + if(!vis[col[i][j]]) ans++; + vis[col[i][j]] = true; + } + } + } + cout << scc - ans << '\n'; + return 0; +} +/* test samples +7 +....... +.##.... +.##.... +....##. +..####. +...###. +....... +*/ \ No newline at end of file diff --git a/test99-4-1 b/test99-4-1 new file mode 100755 index 0000000000000000000000000000000000000000..ead04f4d95b7557f35671078b8f8eee59d6353d4 GIT binary patch literal 49768 zcmeHwdwf*Ywf~++AV7c!qeOXyhl)J&0)ldNc#IH<5R()s(jF$s0Hb+zGQ5P=sd}re z+?Hrs`L#$Bpo&PXS7}RHTCrNor?o92wf%U}2C%(8iq=XkQKq@ywfEXHdrl?<2>tx} z#~j$R_d0v6wbx#It+gNLWahUYpZsTr5DtOMi8KVMq_+_N85u>l#N|lyku0lZ);!-* zU+E&cn|Mj$#9N&p;H*ZjKjQTu(ex0;%|THUN& zzVwKaZfg96S;XTbLh(v;N= 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for(int i = 1; i <= n; i++){ + if(dfs(i, 1)){ + cout << i << '\n'; + break; + } + } + return 0; +} +/* test samples +4 +2 3 4 4 +*/ \ No newline at end of file diff --git a/test99-4-1-2.cpp b/test99-4-1-2.cpp new file mode 100644 index 0000000..9683e58 --- /dev/null +++ b/test99-4-1-2.cpp @@ -0,0 +1,45 @@ +// lanqiao2942 数字王国之军训排队 +#include +using namespace std; +const int N = 15; +int a[N], n; +vector v[N]; + +// 剪枝 +bool dfs(int cnt, int dep){ + if(dep == n+1){ + return true; + } + // 枚举每个人所属的队伍 + for(int i = 1; i <= cnt; i++){ + bool tag = true; + for(const auto &j:v[i]){ + if(a[dep] % j == 0){ + tag = false; + break; + } + } + if(!tag) continue; + v[i].push_back(a[dep]); + if(dfs(cnt, dep + 1)) return true; + // 恢复现场 + v[i].pop_back(); + } + return false; +} +int main(){ + cin >> n; + for(int i = 1; i <= n; i++) cin >> a[i]; + sort(a+1, a+1+n); + for(int i = 1; i <= n; i++){ + if(dfs(i, 1)){ + cout << i << '\n'; + break; + } + } + return 0; +} +/* test samples +4 +2 3 4 4 +*/ \ No 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prefix[l-1] << '\n'; + } + return 0; +} +/* test samples +4 +1 10 +30 50 +60 200 +200 400 +*/ \ No newline at end of file diff --git a/test99-4-3 b/test99-4-3 new file mode 100755 index 0000000000000000000000000000000000000000..d5c80c05f2b20ca05af23c2729c562ec13b42b30 GIT binary patch literal 34840 zcmeI5e{2-T702g(@Y^INxPjnuK2VdkVEo(2lq5a>xPXOY9NVf8Q8S$H*7k<;oqM;2 z1H0j@BqULjwocuKR)w4ik_m#=ia^CtLrp7HO{*%{*GU@65W?KHh&@58w5#5l{*4DUYz-^|# z6P7=>HJ|i!f45uxb|qyn8i`+BaHhXgmcKcc0qN=fY(I7_rPeN6V}TA$^^r-G!5{ccH0-#l-FGlDQha5D`oZi`X+_SXKgCfX340@CkeeK8zN;}^y(^a?5_cH z&ElIH7q+aJp+?Ln#q|zE#g#UYwBxcfmNtf_9hc2e%9i-m-Em`o7Ottzpk?+`K07+1 z5qn~0eD0*OsVl!4pB)R?Wcs`Meb&|1uk+W`+8cEwOU?C>6-v^4iiI;(O@7jSm0dO5ib~Mug*YHWkQddae2T2a z?q#UV^^uP91+0(!?7Q1kCm-})S`Uqx+?UX5W{ARQr%~7*?a*3c+E!C(3kJ>Ls1A0; zVp_x?J#K`8%a<3dEI`nDi@A=9G=I)eC4SBvSJ#F~P21vz)=^MpnP9dMpGACL)D({| z91VoDnCuJ~?NH4^#Sj`#?!@nqNNfC(@GCcp%k025#WOn?b60Vco%m;e)C0!)AjFaajO1egF5 zU;<2l2`~XBzyz286JP>NfC(@GCcp%k025#WOyK{Fzc-;a0yV*iLso*wT$rnm=1 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1e5) return; + if(dep == n + 1){ + cnt[mul]++; + return; + } + // 剪枝2 + int up = pow(1e5 / mul, 1.0/(n - dep + 1)) + 3; + for(int i = st + 1; i < (dep == n?min(sum, up):up); i++){ + dfs(dep+1, i, mul * i, sum + i); + } +} + +int main(){ + int q; cin >> q >> n; + dfs(1, 0, 1, 0); + for(int i = 1; i <= 1e5; i++){ + prefix[i] = prefix[i-1] + cnt[i]; + } + while(q--){ + int l, r; cin >> l >> r; + cout << prefix[r] - prefix[l-1] << '\n'; + } + return 0; +} +/* test samples +4 3 +1 10 +30 50 +60 200 +200 400 +*/ \ No newline at end of file diff --git a/test99-4-4 b/test99-4-4 new file mode 100755 index 0000000000000000000000000000000000000000..484766daa1d16a83241cd51722acb6c887eb634e GIT binary patch literal 34776 zcmeI5e{fXA9l-b99S}8!6rv(RbD>kErTKB>U_y=Z0y#-EO=2#8WHP1eqwIWQwnIaT5?evFfwZsmiGo>?D{86V=CeYTxsMC^|l4Lm2?|0w6 z4}j`NyHE|!=nLxFR{>}WA&Oqua^-)+T12AEJ!jh7|k0Yi~E*Q2`0W38=*W7iT(Ujtsqt=>c01!d7>gRS6V8k`BIDt@VM5($6+iKpbFGcH#sd+}s?LlzA>(D) 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zKljaAx~!sErpFB45->DxbyPRkdwiXrSWzh}{AOuTk75(Q=?%s@Oj*>Tp>TE3Z+bkz zwm_Vl@qlg`)j^-H0c@oZUk~z_VaByUq`HwemT;ecH3wfF>mqGj>(YW9rpDVe{C+pfak0nXfdwZU>4El~EeN%0^|Mm1lKO{Iuao+JrM_M2=cWEz zsgFwi1F4TmeF;nn`}UZ3?27zYI{9H-Xw4=1lkgs&~tye|vdH)epXL@M`sAmnL3&dF%cU zem%bDAD#JaKWjSk>U}4Re;%K@aN(}sZ~M;njpzK&KlJY(eske-kDuImL+?QOSMOe| zy?(Iog~#uF_T;IbeE(|kKX1KrH0Onl$GlU?<*$`?pI`ilZ@m2I&~2?>Y5wRh8 +using namespace std; +using ll = long long; +const ll p = 1e9+7; +const int inf = 1e9, N = 1e5+3; +ll dp[N]; +ll f(int n){ + if( n <= 2) return 1; + if(dp[n] != -1) return dp[n]; + + return dp[n] = (f(n -1 ) + f(n - 2)) % p; +} + +int main(){ + memset(dp, -1, sizeof dp); + int n; cin >> n; + cout << f(n) << endl; + return 0; +} +/* test samples 976496506 +5000 +*/ \ No newline at end of file diff --git a/test99-4-5 b/test99-4-5 new file mode 100755 index 0000000000000000000000000000000000000000..b85776446085331235937b52b122c6aab5d9828b GIT binary patch literal 51688 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zeLYa+TlHr1$Zd&jv5vc6dfTcWwPND_GsQpO{vW;f(Gz?2S3lf+P2GmWCr)|2e?9v5 z{sTK+e!i*py}dh@+<(6D&*cAlSN~hF0`J5J=M}&H>)L5cewf;MOYB#_zxU=}JlAXX I9y&(<1uqUU6aWAK literal 0 HcmV?d00001 diff --git a/test99-4-5.cpp b/test99-4-5.cpp new file mode 100644 index 0000000..2d8e897 --- /dev/null +++ b/test99-4-5.cpp @@ -0,0 +1,66 @@ +#include +using namespace std; +using ll = long long; +const ll p = 1e9 + 7; +const int inf = 1e9, N = 1e3 + 3; + +int n, m, k, sx, sy, fx, fy, h[N][N]; +int dx[] = {0, 0, 1, -1}; +int dy[] = {1, -1, 0, 0}; + +int dp[N][N][2]; + +bool inmp(int x, int y){ + return x >= 1 && x <= n && y >= 1 && y <= m; +} +// 返回值表示能否到达终点(fx, fy), t表示当前使用的喷气背包的次数 +bool dfs(int x, int y, int t){ + if(x == fx && y == fy) return true; + + if(dp[x][y][t] != -1) return dp[x][y][t]; + + for(int i = 0; i < 4; i++){ + int nx = x + dx[i], ny = y + dy[i]; + if(!inmp(nx, ny)) continue; + if(!t){ // 没用过背包 + // 不用 + if(h[x][y] > h[nx][ny] && dfs(nx, ny, 0)) return dp[x][y][t] = true; + // 用 + if(h[x][y] + k > h[nx][ny] && dfs(nx, ny, 1)) return dp[x][y][t] = true; + }else{ // 已经用过一次背包了 + // 不用 + if(h[x][y] > h[nx][ny] && dfs(nx, ny, 1)) return dp[x][y][t] = true; + } + } + return dp[x][y][t] = false; +} +int main(){ + memset(dp, -1, sizeof dp); + cin >> n >> m >> k; + cin >> sx >> sy >> fx >> fy; + for(int i = 1; i <= n; i++){ + for(int j = 1; j <= m; j++){ + cin >> h[i][j]; + } + } + cout << (dfs(sx, sy, 0)?"Yes":"No") << '\n'; + return 0; +} +/* test samples 1 -> YES +5 5 30 +1 1 5 5 +3 20 13 12 11 +19 17 33 72 10 +12 23 12 23 9 +21 43 23 12 2 +21 34 23 12 1 +*/ +/* test samples 1 -> NO +5 5 10 +1 1 5 5 +3 2 13 12 11 +1 17 33 72 10 +12 23 12 23 9 +21 43 23 12 2 +21 34 23 12 1 +*/ \ No newline at end of file