中国剩余定理模数不互质的情况（poj 2891

中国剩余定理模数不互质的情况
主要有一个ax+by==k*gcd(a,b),注意一下倍数情况和最小

https://vjudge.net/problem/POJ-2891

#include <iostream>
#include <cstdio>
#include <queue>
#include <algorithm>
#include <cmath>
#include <cstring>
#define inf 2147483647
#define N 1000010
#define p(a) putchar(a)
#define For(i,a,b) for(long long i=a;i<=b;++i)
//by war
//2019.8.8
using namespace std;
long long T,n;
long long r[N],a[N],x,y,gcd,flag;
void in(long long &x){
    long long y=;char c=getchar();x=;
    while(c<''||c>''){if(c=='-')y=-;c=getchar();}
    while(c<=''&&c>=''){ x=(x<<)+(x<<)+c-'';c=getchar();}
    x*=y;
}
void o(long long x){
    if(x<){p('-');x=-x;}
    if(x>)o(x/);
    p(x%+'');
}

void exgcd(long long a,long long b,long long &x,long long &y){
    if(!b){
        x=;y=;gcd=a;
        return;
    }
    exgcd(b,a%b,y,x);
    y-=a/b*x;
}

signed main(){
    while(scanf("%lld",&n)!=EOF){
        flag=;
        For(i,,n)
           in(r[i]),in(a[i]);
        For(i,,n){
            exgcd(r[],r[i],x,y);
            if((a[i]-a[])%gcd==){
                x*=(a[i]-a[])/gcd;
                y=r[i]/gcd;
                x=(x%y+y)%y;
                a[]+=r[]*x;
                r[]=r[]*r[i]/gcd;
            }
            else{
                o(-);p('\n');
                flag=;
                break;
            }
        }
        if(!flag)
            o(a[]%r[1]),p('\n');
    }
    return ;
}