HDU 3579 Hello Kiki

题解：不互质的中国剩余定理

#include <cstdio>

#include <iostream>

#define ll long long

#define maxn 10

using namespace std;

ll c[maxn],m[maxn],n,am,ac,y0,z0,d;

bool ans;

void exgcd(ll a,ll b,ll& d,ll& x,ll& y){

    if (b==0) { d=a,x=1,y=0; }

    else exgcd(b,a%b,d,y,x),y-=x*(a/b); 

}

int main(){

    ll i,t,cas=0;

    scanf("%I64d",&t);

    while (t--){

        ans=true; scanf("%I64d",&n);

        for(i=1;i<=n;i++)scanf("%I64d",&m[i]);

        for(i=1;i<=n;i++)scanf("%I64d",&c[i]);

        am=m[1]; ac=c[1];

        for (i=2; i<=n; i++){

            exgcd(am,m[i],d,y0,z0);

            if ((ac-c[i])%d!=0){ans=false;break;}

            y0=(c[i]-ac)/d*y0;

            y0=((y0%(m[i]/d))+(m[i]/d))%(m[i]/d);

            ac=am*y0+ac,am=am/d*m[i],ac=(ac%am+am)%am;

        }

        if (ac==0) ac=am;

        printf("Case %I64d: ",++cas);

        if (ans) printf("%I64d\n",ac);

            else printf("-1\n");

    }

    return 0;

}