中国剩余定理模板

互质 情况：

#include <cstdio>//互质
int exGcd(int a,int b,int &x,int &y)
{
    if(b == 0)
    {
        x = 1,y = 0;
        return a;
    }
    int d = exGcd(b,a%b,y,x);
    y -= a/b*x;
    return d;
}
int Chinese_Remainder(int mod[],int prime[],int len)
{
    int i,d,x,y,m,n,ret;
    ret = 0,n = 1;
    for(i=0; i<len; i++) n *= prime[i];
    for(i=0; i<len; i++)
    {
        m = n/prime[i];
        d = exGcd(prime[i],m,x,y);
        ret = (ret+y*m*mod[i])%n;
    }
    return (n+ret%n)%n;
}
int main()
{
    int n,i;
    int mod[15],prime[15];
    while(scanf("%d",&n)&&n)
    {
        for(i=0; i<n; i++)
            scanf("%d%d",&prime[i],&mod[i]);
        printf("%d
",Chinese_Remainder(mod,prime,n));
    }
    return 0;
}

不互质的情况：

//niiick
#include<iostream>
#include<vector>
#include<algorithm>
#include<queue>
#include<cstring>
#include<cstdio>
using namespace std;
typedef long long lt;

lt read()
{
    lt f=1,x=0;
    char ss=getchar();
    while(ss<'0'||ss>'9'){if(ss=='-')f=-1;ss=getchar();}
    while(ss>='0'&&ss<='9'){x=x*10+ss-'0';ss=getchar();}
    return f*x;
}

const int maxn=100010;
int n;
lt ai[maxn],bi[maxn];

lt mul(lt a,lt b,lt mod)
{
    lt res=0;
    while(b>0)
    {
        if(b&1) res=(res+a)%mod;
        a=(a+a)%mod;
        b>>=1;
    }
    return res;
}

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

lt excrt()
{
    lt x,y,k;
    lt M=bi[1],ans=ai[1];//第一个方程的解特判
    for(int i=2;i<=n;i++)
    {
        lt a=M,b=bi[i],c=(ai[i]-ans%b+b)%b;//ax≡c(mod b)
        lt gcd=exgcd(a,b,x,y),bg=b/gcd;
        if(c%gcd!=0) return -1; //判断是否无解，然而这题其实不用
        
        x=mul(x,c/gcd,bg);
        ans+=x*M;//更新前k个方程组的答案
        M*=bg;//M为前k个m的lcm
        ans=(ans%M+M)%M;
    }
    return (ans%M+M)%M;
}

int main()
{
    n=read();
    for(int i=1;i<=n;++i)
    bi[i]=read(),ai[i]=read();
    printf("%lld",excrt());
    return 0;
}