POJ2891——Strange Way to Express Integers(模线性方程组)

Strange Way to Express Integers

DescriptionElina is reading a book written by Rujia Liu, which introduces a strange way to express non-negative integers. The way is described as following:
Choose k different positive integers a1, a2, …, ak. For some non-negative m, divide it by every ai (1 ≤ i ≤ k) to find the remainder ri. If a1, a2, …, ak are properly chosen, m can be determined, then the pairs (ai, ri) can be used to express m.
“It is easy to calculate the pairs from m, ” said Elina. “But how can I find m from the pairs?”
Since Elina is new to programming, this problem is too difficult for her. Can you help her?
Input
The input contains multiple test cases. Each test cases consists of some lines.
    Line 1: Contains the integer k.
    Lines 2 ~ k + 1: Each contains a pair of integers ai, ri (1 ≤ i ≤ k).
Output
Output the non-negative integer m on a separate line for each test case. If there are multiple possible values, output the smallest one. If there are no possible values, output -1.
Sample Input
2
8 7
11 9
Sample Output
31

题目大意：

　　　　给定T个（a，b） 求解最小整数 X，使X满足 X mod a = b

解题思路：

　　　　http://www.cnblogs.com/Enumz/p/4063477.html

Code：

/*************************************************************************
    > File Name: poj2891.cpp
    > Author: Enumz
    > Mail: 369372123@qq.com
    > Created Time: 2014年10月28日 星期二 02时50分07秒
 ************************************************************************/

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<string>
#include<cstring>
#include<list>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<algorithm>
#include<cmath>
#include<bitset>
#include<climits>
#define MAXN 100000
#define LL long long
using namespace std;
LL extended_gcd(LL a,LL b,LL &x,LL &y) //返回值为gcd(a,b)
{
    LL ret,tmp;
    if (b==0)
    {
        x=1,y=0;
        return a;
    }
    ret=extended_gcd(b,a%b,x,y);
    tmp=x;
    x=y;
    y=tmp-a/b*y;
    return ret;
}
int main()
{
    LL N;
    while (cin>>N)
    {
        long long a1,m1;
        long long a2,m2;
        cin>>a1>>m1;
        if (N==1)
            printf("%lld\n",m1);
        else
        {
            bool flag=0;
            for (int i=2;i<=N;i++)
            {
                cin>>a2>>m2;
                if (flag==1) continue;
                long long x,y;
                LL ret=extended_gcd(a1,a2,x,y);
                if ((m2-m1)%ret!=0)
                    flag=1;
                else
                {
                    long long ans1=(m2-m1)/ret*x;
                    ans1=ans1%(a2/ret);
                    if (ans1<0) ans1+=(a2/ret);
                    m1=ans1*a1+m1;
                    a1=a1*a2/ret;
                }
            }
            if (!flag)
                cout<<m1<<endl;
            else
                cout<<-1<<endl;
        }
    }
    return 0;
}