【洛谷P4777】扩展中国剩余定理（Excrt）

Description

给定若干个形如$xequiv a_i pmod {b_i}$的同余方程，其中b不保证两两互质，求最小非负整数解x

Solution

扩展中国剩余定理的模板题。

假定我们已经求出了前k-1个方程的一个解为x

并且记$M=prodlimits_{i=1}^{k-1}{b_i}$

那么前k-1个方程的通解为$x+iM$

考虑当前第k个方程，我们要求出一个t，使得$x+tMequiv a_k pmod {b_k}$

变形，得$tM+Nb_k=a_k-x$

这个式子可以用Exgcd求解，若Exgcd有解，那么满足前k个方程的一个解为$x_k=x+tM$

所以我们求解n次扩展欧几里得算法即可得到解，时间复杂度为$O(nlogn)$

Code

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
inline ll read() {
    ll 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 mul(ll a, ll b, ll mod) {
    ll ret = 0;
    while (b) {
        if (b & 1) ret = ret + a % mod;
        a = a + a % mod;
        b >>= 1;
    }
    return ret;
}
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 ans, a[100010], b[100010], n, M;
inline ll excrt() {
    ans = a[1]; M = b[1];
    for (register int i = 2; i <= n; ++i) {
        ll x, y, A = M, B = b[i], C = (a[i] - ans % b[i] + b[i]) % b[i];
        ll gcd = exgcd(A, B, x, y);
        x = mul(x, C / gcd, B / gcd);
        ans += x * M;
        M *= B / gcd;
        ans = (ans % M + M) % M;
    }
    return (ans % M + M) % M;
}
int main() {
    n = read();
    for (register int i = 1; i <= n; ++i) {
        b[i] = read(), a[i] = read();
    }
    printf("%lld
", excrt());
    return 0;
}