POJ2891 Strange Way to Express Integers(解多元线性同余方程&&非互质情况)

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, …, 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

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2
8 7
11 9

Sample Output

31

给出n个等式关系，但与中国剩余定理不同的是它不是互质的情况

它需要不断合并方程，最终合并成一个

比如上面的例子

m%8==7

m%11=9

转化成m=k*8+7

m=k*11+9

也就是x=k1*a1+r1,x=k2*a2+r2

联立

a1*k1-a2*k2=(r2-r1)。

#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;//判断
            }
            int t = b / d;
            x0 = (x0 * (c / d) % t + t) % t;
            r1 = a1 * x0 + r1;//更新方程组值
            a1 = a1 * (a2 / d);
        }
        if(!ifhave)
        {
            printf("-1\n");
            continue;
        }
        printf("%lld\n",r1);
    }
    return 0;
}