poj 2891(中国剩余定理)

Strange Way to Express Integers

Time Limit: 1000MS		Memory Limit: 131072K
Total Submissions: 13056		Accepted: 4167

Description

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?

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 a_i, r_i (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

Hint

All integers in the input and the output are non-negative and can be represented by 64-bit integral types.

直接用kuangbin大神的模板了，话说kuangbin大神的模板真好用，，无论模的数是不是互质都可以用。。

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <math.h>
using namespace std;
typedef long long LL;
const int N = 1000;
LL extend_gcd(LL a,LL b,LL &x,LL &y)
{
    if(b==0)
    {
        x=1,y=0;
        return a;
    }
    else
    {
        LL x1,y1;
        LL d = extend_gcd(b,a%b,x1,y1);
        x = y1;
        y = x1-a/b*y1;
        return d;
    }
}
LL m[N],a[N];///模数为m,余数为a, X % m = a
bool solve(LL &m0,LL &a0,LL m,LL a)
{
    long long y,x;
    LL g = extend_gcd(m0,m,x,y);
    LL t = a-a0>0?a-a0:a0-a;
    if( t%g )return false;
    x *= (a - a0)/g;
    x %= m/g;
    a0 = (x*m0 + a0);
    m0 *= m/g;
    a0 %= m0;
    if( a0 < 0 )a0 += m0;
    return true;
}
/**
* 无解返回false,有解返回true;
* 解的形式最后为 a0 + m0 * t (0<=a0<m0)
*/
bool MLES(LL &m0 ,LL &a0,LL n)///解为 X = a0 + m0 * k
{
    bool flag = true;
    m0 = 1;
    a0 = 0;
    for(int i = 0; i < n; i++)
        if( !solve(m0,a0,m[i],a[i]) )
        {
            flag = false;
            break;
        }
    return flag;
}
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        for(int i=0; i<n; i++)
        {
            scanf("%lld%lld",&m[i],&a[i]);
        }
        LL m0,a0;
        bool flag = MLES(m0,a0,n);
        if(!flag) printf("-1
");
        else
        {
            LL x = (a0%m0+m0)%m0;
            printf("%lld
",x);
        }
    }
    return 0;
}