• 一个整数除以3余2、除以5余3、除以7余2，求这个整数？答案：23

  • 构造逆元方程 ax + np = 1

  • 获得逆元 x

  • 求权重和

  • 求余和正数处理

代码

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
LL ex_gcd(LL a, LL b, LL &x, LL &y)
{if (b == 0){x = 1;y = 0;return a;}LL gcd = ex_gcd(b, a % b, x, y);LL tmp = x;x = y;y = tmp - a / b * y;return gcd;
}
int main() // ax + np = 1
{LL mod3, mod5, mod7;cin >> mod3 >> mod5 >> mod7;LL yy1, yy2, yy3; // a[] = 35, 21, 15  p[] = 3, 5, 7LL x[4], n[4];LL gcd1 = ex_gcd(35, 3, x[1], n[1]);LL gcd2 = ex_gcd(21, 5, x[2], n[2]);LL gcd3 = ex_gcd(15, 7, x[3], n[3]);if (gcd1 != 1 || gcd2 != 1 || gcd3 != 1){cout << "No solution" << endl;return 0;}yy1 = 35 * x[1];yy2 = 21 * x[2];yy3 = 15 * x[3];LL ans = mod3 * yy1 + mod5 * yy2 + mod7 * yy3;LL mod = 3 * 5 * 7;cout << (ans % mod + mod) % mod;return 0;
}