poj 2891（中国剩余定理）

Strange Way to Express Integers

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

题意：给你k组数据  接下来是ai  ri  问你是否存在一个数x  满足x%ai=ri  若存在输出最小的一个  否则输出-1；

分析：

先考虑k==2的情况：

x = a1 ( mod m1 )

x = a2 ( mod m2 )

方程组有解的充分必要条件是： d | (a1-a2) ，其中 d = (m1,m2)

证明如下：

必要性： 设 x 是上面同余方程组的解，从而存在整数q1,q2使得x=a1+m1*q1，x=a2+m2*q2，消去x即得a1-a2 = m2q2-m1q1。由于d=(m1,m2)，故d | (a1-a2)。

充分性：若d=(m1,m2) | (a1-a2)成立，则方程m1*x + m2*y = a1-a2有解。

设解为x0,y0。那么m2*y0 = a1-a2 ( mod m1 )

记x1 = a2+m2*y0，可以知道 x1=a2 ( mod m2 )，且x1 = a2+m2*y0 = a2 + ( a1-a2) = a1 ( mod m1 )

所以 x1 = a2 ( mod m2 ) = a1 ( mod m1 )

所以 x = x1 ( mod [m1,m2] )

另外，若x1与x2都是上面同余方程组的解，则 x1 = x2 ( mod m1 )， x1 = x2 ( mod m2 )， 由同余的性质得 x1 = x2 ( mod [m1,m2] )，即对于模[m1,m2]，同余方程组的解释唯一的。

PS:由于ai，aj不保证互素，不能用直接套中国剩余定理，做法是利用欧几里德扩展定理，将两个等式合并，然后再与其他的等式一一合并

更详细的解释请参考算法导论p556

#include <queue>
#include <stack>
#include <math.h>
#include <vector>
#include <limits.h>
#include <stdio.h>
#include <iostream>
#include <string.h>
#include <algorithm>
#include <functional>
#define N 100010
#define mem(a) memset(a,0,sizeof(a));
#define mem_1(a) memset(a,-1,sizeof(a));
#define LL long long
using namespace std;
LL k,m,t;
LL gcd(LL a,LL b)
{
    if(b==0)
        return a;
    else return (b,a%b);
}
LL extend(LL a,LL b,LL &x,LL &y)
{
    if(b==0)
    {
        x=1;
        y=0;
        return a;
    }
    LL d=extend(b,a%b,x,y);
    t=x;
    x=y;
    y=t-a/b*y;
    return d;
}
int main()
{
    int i,flag;
    LL p,q,d,a1,a2,b1,b2,x,c;
    while(scanf("%lld",&k)!=EOF)
    {
        scanf("%lld%lld",&b1,&a1);
        flag=0;
        for(i=0; i<k-1; i++)
        {
            scanf("%lld%lld",&b2,&a2);
            if(flag)
                continue;
            d=extend(b1,b2,p,q);
            c=a2-a1;
            if(c%d)
            {
                flag=1;
                continue;
            }
            t=b2/d;
            x=(c/d*p%b2+b2)%t;
            a1=x*b1+a1;
            b1=b1*b2/d;
        }
        if(flag)
            puts("-1");
        else
            printf("%lld\n",a1);
    }
    return 0;
}