poj 2981 Strange Way to Express Integers (中国剩余定理不互质)

http://poj.org/problem?id=2891

Strange Way to Express Integers

Time Limit: 1000MS		Memory Limit: 131072K
Total Submissions: 11970		Accepted: 3788

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

题目大意： x % ai = ri 求满足条件的最小的x

刚开始看中国剩余定理，直接套用中国剩余定理模板，结果各种RE，原来还有不是两两互质的情况，还是so young 啊！！！！

那么应该怎么处理这种情况呢， 合并方程求解

#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<stdlib.h>

using namespace std;

const int N = 1010;
typedef __int64 ll;
ll r, n[N], b[N];

void gcd(ll a, ll b, ll &x, ll &y)
{
    if(b == 0)
    {
        x = 1;
        y = 0;
        r = a;
        return ;
    }
    gcd(b, a % b, x, y);
    ll t = x;
    x = y;
    y = t - a / b * y;
}

ll CRT2(ll n[], ll b[], ll m)
{
    int f = 0;
    ll n1 = n[0], n2, b1 = b[0], b2, c, t, k, x, y;
    for(ll i = 1 ; i < m ; i++)
    {
        n2 = n[i];
        b2 = b[i];
        c = b2 - b1;
        gcd(n1, n2, x, y);//扩展欧几里德
        if(c % r != 0)//无解
        {
            f = 1;
            break;
        }
        k = c / r * x;//扩展欧几里德求得k
        t = n2 / r;
        k = (k % t + t) % t;
        b1 = b1 + n1 * k;
        n1 = n1 * t;
    }
    if(f == 1)
        return -1;
    return b1;
}

int main()
{
    ll k;
    while(~scanf("%I64d", &k))
    {
        for(ll i = 0 ; i < k ; i++)
            scanf("%I64d%I64d", &n[i], &b[i]);
        printf("%I64d
", CRT2(n, b, k));
    }
    return 0;
}