51Nod 1079 Chinese remainder theorem template [Title]

 

#include<stdio.h> 
#define ll long long
/*
    gcd(a,b) = Xa+Yb
    Xb+Y(a%b) = Xb + Y( a - a/b*b ) = Xb + Ya - a/b*Y*b = Ya + (X-a/b*Y)b

*/
ll x,y;
void exgcd(ll a,ll b){
    if( a%b==0 ){
        x = 0;
        y = 1;
    }
    else{
        exgcd(b,a%b);
        ll temp = x;
        x = y;
        y = temp - a/b*y;
    }
}
ll N=1,ans;
int k[15],p[15];

int main(){
    
    int n; scanf("%d",&n);
    for(int i=1;i<=n;i++){
        scanf("%d%d",p+i,k+i);
        N*=p[i];
    }
    for(int i=1;i<=n;i++){
        /*p[i] 与 N/p[i] 互质
        gcd(p[i],N/p[i]) = 1
        X*p[i] + Y*N/p[i] = 1
        Y是N/p[i]模p[i]下的乘法逆元
        */
        exgcd(p[i],N/p[i]);
        //cout<<x<<" "<<p[i]<<" "<<y<<" "<<N/p[i]<<endl;

        ans = (ans + k[i]*N/p[i]*y)%N;
    }
    printf("%lld",(ans % N + N) % N);

}