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 

如果flag==0  说明无解

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
typedef long long ll;
bool flag;
ll m[100],a[100],lcm;
ll gcd(ll a,ll b){
    return b ? gcd(b,a%b) : a;
}
ll ex_gcd(ll a,ll b,ll &x,ll &y){
    if(!b){
        x = 1;
        y = 0;
        return a;
    }
    ll g = ex_gcd(b,a%b,y,x);
    y -= a / b * x;
    return g;
}
ll China(ll n){
    ll m1 = m[0],a1 = a[0];
    ll m2,a2,k1,k2,x0,g,c;
    lcm = m[0];
    for(int i = 1; i < n; i++){
        m2 = m[i];
        a2 = a[i];
        c = a2 - a1;
        g = ex_gcd(m1,m2,k1,k2);
        lcm = lcm * m[i] / gcd(lcm,m[i]);
        if(c % g){
            flag = false;
            return 0;
        }
        x0 = k1 * c / g;
        ll t = m2 / g;
        x0 = (x0 % t + t) % t;
        a1 += m1 * x0;
        m1 = m2 / g * m1;
    }
    return a1;
}
int main(){
    int T;
    scanf("%d",&T);
    while(T--){
        ll N;
        int n;
        scanf("%lld%d",&N,&n);
        for(int i = 0; i < n; i++) scanf("%lld",&m[i]);
        for(int i = 0; i < n; i++) scanf("%lld",&a[i]);
        flag = true;
        ll ans = China(n);
        if(!flag || ans > N){
            printf("0
");
        }
        else{
            if(ans <= 0) ans += lcm;
            ans = (N - ans) / lcm + 1;
            printf("%lld
",ans);
        }
    }
    return 0;
}