[模板] 中国剩余定理

[证明] 待补充


//Writer:GhostCai && His Yellow Duck

#include<iostream>
using namespace std;

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

int china(int *M,int *B,int k){
    int x,b,a=0,m=1,mi;
    for(int i=1;i<=k;i++) m*=M[i];
    for(int i=1;i<=k;i++){
        mi=m/M[i];
        exgcd(M[i],mi,x,b);
        a=(a+b*mi*B[i])%m;
    }
    if(a>0) return a;
    else return a+m;
}

int main(){
    int b[100],m[100];
    int n;
    cin>>n;
    for(int i=1;i<=n;i++){
        cin>>b[i]>>m[i];
    }
    cout<<china(m,b,n);
}

本文来自博客园,作者:GhostCai,转载请注明原文链接:https://www.cnblogs.com/ghostcai/p/9247515.html

原文地址:https://www.cnblogs.com/ghostcai/p/9247515.html