POJ2891

求解线性同余方程组，模数可能不两两互质，因此需要使用合并线性同余方程的方法。

#include <stdio.h>

typedef long long int llt;

//The extended Euclidean algorithm implemented by iteration
//returns gcd(a,b), and x, y are satisfied with ax + by = gcd
llt exEuclid(llt a,llt b,llt&x,llt&y){
    llt x0 = 1, y0 = 0;
    llt x1 = 0, y1 = 1;
    x = 0; y = 1;
    llt r = a % b;
    llt q = ( a - r ) / b;
    while( r ){
        x = x0 - q * x1;
        y = y0 - q * y1;
        x0 = x1; y0 = y1;
        x1 = x; y1 = y;
        a = b; b = r; r = a % b;
        q = ( a - r ) / b;
    }
    return b;
}

//to convert two congruence equations to equivalent one
//x = r1 (mod m1)
//x = r2 (mod m2)
//the out put is:  x = r3 (mod m3)
//return true if there is a solution, otherwise false
bool mergeCrt(llt r1,llt m1,llt r2,llt m2,llt&r3,llt&m3){
    llt x,y;
    llt g = exEuclid(m1,m2,x,y);
    llt r = (r2 - r1) % m2;
    if ( r < 0 ) r += m2;

    if ( r % g ) return false;//no solution

    x = r / g * x % ( m2 / g );//the least positive solution

    m3 = m1 / g * m2; //lcm
    r3 = ( ( x * m1 ) % m3 + r1 ) % m3;
    return true;
}

int main(){
    int n;
    while ( EOF != scanf("%d",&n) ){
        llt m1=1,r1=0,m2,r2;
        bool flag = true;

        for(int i=0;i<n;++i){
            scanf("%lld%lld",&m2,&r2);
            if ( flag ){
                flag = mergeCrt(r1,m1,r2,m2,r1,m1);
            }
        }

        r1 %= m1;
        if ( r1 < 0 ) r1 += m1;
        printf("%lld\n",flag?r1:-1LL);
    }
    return 0;
}