POJ2891 Strange Way to Express Integers(解多元线性同余方程)

Strange Way to Express Integers

Time Limit: 1000MS		Memory Limit: 131072K
Total Submissions: 11289		Accepted: 3482

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.

题目大意:

选择a1,a2....ak,对于某个整数m分别对ai求余对应整数ri,现在已知a1,a2,....ak,以及整数对(ai,ri),求非负整数m的值,若有多个m，输出最小的一个。

解题思路:

容易列出方程 m % ai = ri,可以转化为同余方程 m ≡ ri (mod ai),接着套用求解多元线性同余方程的模版即可。

AC代码:

#include<iostream>
#include<cstdio>
#include<iostream>

using namespace std;

typedef long long ll;

void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y)
{
    if(!b)
    {
        x = 1,y = 0,d = a;
    }
    else
    {
        ex_gcd(b,a%b,d,y,x);
        y -= x * (a / b);
    }
}

int main()
{
    long long i,n,a1,r1,a2,r2,a,b,c,d,x0,y0;
    while(scanf("%lld",&n) != EOF)
    {
        bool ifhave = 1;
        scanf("%lld%lld",&a1,&r1);
        for(i=1;i<n;i++)
        {
            scanf("%lld%lld",&a2,&r2);
            a = a1,b = a2,c = r2 - r1;
            ex_gcd(a,b,d,x0,y0);
            if(c % d != 0)
            {
                ifhave = 0;
            }
            int t = b / d;
            x0 = (x0 * (c / d) % t + t) % t;
            r1 = a1 * x0 + r1;
            a1 = a1 * (a2 / d);
        }
        if(!ifhave)
        {
            printf("-1\n");
            continue;
        }
        printf("%lld\n",r1);
    }
    return 0;
}