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中国剩余定理的板子。小心取模。

#include <bits/stdc++.h>
using namespace std;

const int N = 11;
#define int long long


int n, M = 1, a[N], b[N], t[N], ans;

int fmul (int x, int y, int mod) {
    int res = 0;
    while (y) {
        if (y & 1) {
            res = (res + x) % mod;
        }
        x = (x + x) % mod;
        y >>= 1;
    }
    return res;
}

void exgcd (int a, int b, int &x, int &y) {
    if (b == 0) {x = 1, y = 0; return;}
    exgcd (b, a % b, x, y);
    int xx = y, yy = x - (a / b) * y;
    x = xx, y = yy;
}

signed main () {
    cin >> n;
    for (int i = 1; i <= n; ++i) cin >> a[i];
    for (int i = 1; i <= n; ++i) cin >> b[i];
    for (int i = 1; i <= n; ++i) a[i] = ((a[i] % b[i]) + b[i]) % b[i];
    for (int i = 1; i <= n; ++i) M *= b[i];
    for (int i = 1; i <= n; ++i) {
        exgcd (M / b[i], b[i], t[i], t[0]);
        t[i] = ((t[i] % b[i]) + b[i]) % b[i];
        ans = (ans + fmul (fmul (M / b[i], t[i], M), a[i], M)) % M; 
    }
    cout << ((ans % M) + M) % M << endl;
    
}