扩展中国剩余定理

http://blog.youkuaiyun.com/litble/article/details/75807726

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#define ll long long
using namespace std;
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while (ch<'0'||ch>'9'){if (ch=='-') f=-1;ch=getchar();}
    while (ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int n;
ll m[100005],a[100005],x,y;
ll ex_gcd(ll a,ll b)
{
    if (!b){x=1;y=0;return a;}
    ll d=ex_gcd(b,a%b),t=x;
    x=y;y=t-(a/b)*y;
    return d;
}
inline ll solve()
{
    ll M=m[1],A=a[1],t;
    for (int i=2;i<=n;i++)
    {
        ll d=ex_gcd(M,m[i]);
        if ((a[i]-A)%d) return -1;
        x*=(a[i]-A)/d;t=m[i]/d;x=(x%t+t)%t;
        A=M*x+A;M=M/d*m[i];A%=M;
    }
    return (A%M+M)%M;
}
int main()
{
    while (scanf("%d",&n)!=EOF)
    {
        for (int i=1;i<=n;i++) scanf("%lld%lld",&m[i],&a[i]);
        printf("%lld\n",solve());
    }
    return 0;
}