POJ 2891-trange Way to Express Integers（解线性同余方程组）-蒲公英云

POJ 2891-trange Way to Express Integers（解线性同余方程组）

Strange Way to Express Integers

Time Limit: 1000MS		Memory Limit: 131072K
Total Submissions: 13819		Accepted: 4451

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, …, 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

Hint

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

Source

POJ Monthly—2006.07.30, Static

题目意思：

有多组测试数据，每组先给出一个N，表示该组有N个线性同余方程。

对每一组测试数据，求解一个M满足： M≡Ri（mod Ai）

解题思路：

模板题，线性同余方程组套一个拓展欧几里得。

/*
* Copyright (c) 2016, 烟台大学计算机与控制工程学院
* All rights reserved.
* 文件名称：exgcd.cpp
* 作    者：单昕昕
* 完成日期：2016年4月2日
* 版 本 号：v1.0
*/
#include<iostream>
#include<cstring>
#include<cstdio>
#include<malloc.h>
using namespace std;
typedef long long ll;
ll exgcd(ll a,ll b,ll& x,ll& y)//拓展欧几里得
{
    if(b==0)
    {
        x=1;
        y=0;
        return a;  //d=a，x=1,y=0,此时等式d=ax+by成立
    }
    ll d=exgcd(b,a%b,y,x);
    y-=x*(a/b); //系数x、y的取值是为满足等式d=ax+by
    return d;
}
ll solve(ll n)//解同余方程组
{
    ll a1,a2,r1,r2,x0,y0;
    bool ifhave=true;
    cin>>a1>>r1;
    for(int i=1; i<n; ++i)
    {
        cin>>a2>>r2;
        ll a=a1,b=a2,c=r2-r1;
        ll d=exgcd(a,b,x0,y0);
        if(c%d!=0) ifhave=false;
        int t=b/d;
        x0=(x0*(c/d)%t+t)%t;
        r1=a1*x0+r1;
        a1=a1*(a2/d);
    }
    if(!ifhave) r1=-1;
    return r1;
}
int main()
{
    ll t;
    while(cin>>t)
    {
        cout<<solve(t)<<endl;
    }
    return 0;
}
/**
2
8 7
11 9
**/