【洛谷P1495】曹冲养猪

Description

给定若干个形如$xequiv a_i pmod {m_i}$的同余方程，其中m两两互质，求x

Solution

这是中国剩余定理的模板题，具体解法如下：

我们首先求出$M=prodlimits_{i=1}^{n}{m_i}$

那么令$M_i=frac{M}{m_i}$

然后令$t_i=M_i^{-1} pmod {m_i}$

最后令$s_i=t_iM_i$

我们只要求出$xequiv sumlimits_{i=1}^{n}{a_is_i}pmod M$即可

逆元可以用Exgcd求解，时间复杂度为O(nlogn)

Code

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
inline int read() {
    int ret = 0, op = 1;
    char c = getchar();
    while (!isdigit(c)) {
        if (c == '-') op = -1; 
        c = getchar();
    }
    while (isdigit(c)) {
        ret = ret * 10 + c - '0';
        c = getchar();
    }
    return ret * op;
}
ll exgcd(ll a, ll b, ll &x, ll &y) {
    if (!b) {
        x = 1, y = 0;
        return a;
    }
    ll gcd = exgcd(b, a % b, x, y);
    ll x2 = x, y2 = y;
    x = y2;
    y = x2 - (a / b) * y2;
    return gcd;
}
ll n, a[12], b[12], mod = 1;
ll ans;
int main() {
    n = read();
    for (register int i = 1; i <= n; ++i) {
        a[i] = read(), b[i] = read();
        mod *= a[i];
    }
    for (register int i = 1; i <= n; ++i) {
        ll m = mod / a[i];
        ll x = 0, y = 0;
        exgcd(m, a[i], x, y);
        ans = (ans + (x * m * b[i]) % mod + mod) % mod;
    }
    printf("%lld
", ans);
    return 0;
}