poj 2891 解题报告

//中国剩余定理  因子非互素,所以只能两个两个得来；
// 证明：
//x = a1(mod n1);
//x = a2(mod n2);
//因为 n1- n2 不互素;
//so : x = n1* k1 + a1;
//　　x = n2 *k2 + a2;
//两式相减
//n1*k1 = n2*k2 + a2-a1;
//n1 *k1 = a2-a1(mod n2);
//若k1 有解，则gcd(n1,n2)|(a2-a1);
//设 d = gcd(n1,n2);
//n1*k1/d = (a2-a1)/d (mod n2/d);
//k1 = (a2-a1)/d *  (n1/d)^-1 (mod n2/d);

//则：x = n1*((a2-a1)/d *(n1/d)^-1 )+k(n2/d) )+ a1;

//x=  n1*(a2-a1)*(n1/d)^-1 /d + a1(mod n1n2/d);

//令 t =n1*(a2-a1)*(n1/d)^-1 ，n =  n1n2/d;

//x = tn1 + a2(mod n);
//所以所有方程可以合并成一个。。。

#include"iostream"
using namespace std;

long long gcd(long long a,long long b,long long &x,long long &y  ){
    if(b==0){
        x = 1;
        y = 0;
        return a;
    }
    long long ans = gcd(b,a%b,x,y);
    long long t = x;
    x = y;
    y = t- a/b*y;
    return ans;
}
int main(){
    long long n;
    while(cin>>n){
        long long m1,r1,m2,r2,a,b,x,y;
        int flag = 1,i;
        cin>>m1>>r1;
        for(i=1;i<n;i++){
            cin>>m2>>r2;
            if(!flag)
                continue;
            a = gcd(m1,m2,x,y);
            if((r2-r1)%a!=0){
                flag = 0;
                continue;
            }
            x = x*(r2-r1)/a;
            b = m2/a;
            x= (x%b+b)%b;
            r1 = r1 + m1*x;
            m1 = (m1*m2)/a;
        }
        if(flag)
            cout<<r1<<endl;
        else
            cout<<-1<<endl;
    }
    return 0;
}