题目链接：
POJ 2891 Strange Way to Express Integers
题意:
给出x模m[i]的余数a[i]，求x。
分析：
中国剩余定理。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <climits>
#include <cmath>
#include <ctime>
#include <cassert>
#define IOS ios_base::sync_with_stdio(0); cin.tie(0);
using namespace std;
typedef long long ll;
const int MAX_M = 100100;int M;
ll a0, m0;
ll a[MAX_M], m[MAX_M];ll ex_gcd(ll a, ll b, ll& x, ll& y)
{if(b == 0){x = 1, y = 0;return a;}ll dd = ex_gcd(b, a % b, y, x);y -= a / b * x;return dd;
}bool ModularLinearEquation(ll& m0, ll& a0, ll mm, ll aa)
{ll x, y, d;d = ex_gcd(m0, mm, x, y);if(labs(aa - a0) % d) return false;mm /= d;x = x * (aa - a0) / d % mm;a0 += x * m0;m0 *= mm;a0 = (a0 % m0 + m0) % m0;return true;
}bool CRT()
{bool flag = true;m0 = 1, a0 = 0;for(int i = 0; i < M; i++){if(ModularLinearEquation(m0, a0, m[i], a[i]) == false){flag = false;break;}}return flag;
}int main()
{while(~scanf("%d", &M)){for(int i = 0; i < M; i++){scanf("%lld%lld", &m[i], &a[i]);}if(CRT() == false) printf("-1\n");else printf("%lld\n", a0);} return 0;
}