Strange Way to Express Integers

Time Limit: 1000MS		Memory Limit: 131072K
Total Submissions: 8005		Accepted: 2378

Elina 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, …, a_k. For some non-negative m, divide it by every a_i (1 ≤ i ≤ k) to find the remainder r_i. If a₁, a₂, …, a_k are properly chosen, m can be determined, then the pairs (a_i, r_i) 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?

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 a_i, r_i (1 ≤ i ≤ k).

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.

2

8 7

11 9

31
题目大意：k个模线性方程求解
AC代码：

#include <iostream>

using namespace std;

__int64 Extended_Euclid(__int64 a,__int64 b,__int64 &x,__int64 &y)

{

    __int64 t,d;

    if(b==)

    {x=;y=;return a;}

    d = Extended_Euclid(b,a%b,x,y);

    t = x;

    x=y;

    y=t-(a/b)*y;

    return d;

}

__int64 CRT(__int64 n)

{

    __int64 i,x,y;

    __int64 d,c,t,m,b,mm,bb;

    int flag=;

    scanf("%I64d %I64d",&m,&b);

    mm=m;bb=b;

    if(n==)

        return b%m;

    for(i=;i<n;i++)

    {

        scanf("%I64d %I64d",&m,&b);

        if(flag)

        {

            d=Extended_Euclid(mm,m,x,y);

            c=b-bb;

            if(c%d)

                flag=;

            else

            {

                t=m/d;

                bb=mm*((c/d*x%t+t)%t)+bb;

                mm=mm*m/d;

            }

        }

    }

    if(flag)

        return bb%mm;

    else

        return -;

}

int  main()

{

    int n;

    while(cin>>n)

        printf("%I64d\n",CRT(n));

    return ;

}