28 lines
657 B
C++
28 lines
657 B
C++
// lanqiao 1457 杨辉三角形(朴素解法)
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#include<bits/stdc++.h>
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using namespace std;
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#define int long long
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const int N = 1e3 + 10;
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int a[N][N];
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signed main(){
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int n; cin >>n;
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if(n == 1){
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cout << 1 << endl;
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return 0;
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}
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a[1][1] = a[2][1] = a[2][2] = 1;
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for(int i = 3; i <= N-1; i++){
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for(int j = 1; j <= i; j++){
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if(j == 1 || j == i) a[i][j] = 1;
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else a[i][j] = a[i-1][j] + a[i-1][j-1];
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if(a[i][j] == n){ // 1 + 2 + ... + i-1
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cout << i*(i-1)/2 + j << endl;
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return 0;
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}
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}
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}
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return 0;
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}
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/* test samples -> 13
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6
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*/ |