From 755a358c75c471ba1a01c0589b58a283d840975a Mon Sep 17 00:00:00 2001
From: xingyou wu <3050128610@qq.com>
Date: Mon, 31 Mar 2025 19:48:16 +0800
Subject: [PATCH] =?UTF-8?q?lanqiao=202108=20=20X=20=E8=BF=9B=E5=88=B6?=
 =?UTF-8?q?=E5=87=8F=E6=B3=95?=
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---
 13lanqiao/test5.cpp | 34 ++++++++++++++++++++++++++++++++++
 1 file changed, 34 insertions(+)
 create mode 100644 13lanqiao/test5.cpp

diff --git a/13lanqiao/test5.cpp b/13lanqiao/test5.cpp
new file mode 100644
index 0000000..aa0ff96
--- /dev/null
+++ b/13lanqiao/test5.cpp
@@ -0,0 +1,34 @@
+// lanqiao 2108  X 进制减法
+#include<bits/stdc++.h>
+using namespace std;
+#define int long long
+const int N = 1e5+10, MOD = 1e9+7;
+int a[N], b[N];
+/*
+解法: 通过观察可得，每一位的进制=max(max(a[i],b[i])+1,2)
+同时为了保证个位对齐，所以需要将a,b数组逆序存储
+*/
+signed main(){
+    int mx, m, n; // mx-最高进制位, A的进制位m, B的进制位n
+    cin >> mx >> m;
+    for(int i = m; i >= 1; i--) cin >> a[i];
+    cin >> n;
+    for(int i = n; i >= 1; i--) cin >> b[i];
+    int ans = 0; // 存储A-B的结果
+    int w = 1; // W存储每一位的权重
+    int t; // 存储每一位的进制
+    for(int i = 1; i <= max(m,n); i++){
+        t = max(max(a[i], b[i]) + (int)1, (int)2); // 计算每一位的进制
+        ans = (ans + (a[i] - b[i]) * w) % MOD; // 计算当前位置减法后转化为十进制的结果
+        w = (w*t) % MOD; // 计算每一位的权重
+    }
+    cout << ans % MOD << endl;
+    return 0;
+}
+/* test samples -> 94
+11
+3
+10 4 0
+3
+1 2 0
+*/
\ No newline at end of file