pog 2891 Strange Way to Express Integers (中国剩余定理 非互质)

Strange Way to Express Integers

Time Limit: 1000MS Memory Limit: 131072KTotal Submissions: 14789 Accepted: 4847

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

28 711 9

Sample Output

31

题解：题意就是线性方程组的解 x;

x = r1 (mod a1);

x = r2 (mod a2);

……

x = rn (mod an).

a1,a2……an,不两两互质的情况

两两合并求解

自己不能直接写下来，只能记模板，使用逆元还是有点不理解！

代码：

#include<cstdio>#define LL __int64LL gcd(LL a,LL b){    if(b==0)        return a;    return gcd(b,a%b);}LL exgcd(LL a,LL b,LL &x,LL &y){    if(b==0)    {        x=1;        y=0;        return a;    }    LL d,temp;    d=exgcd(b,a%b,x,y);    temp=x;    x=y;    y=temp-a/b*y;    return d;}int main(){    LL k,a1,a2,r1,r2,c,d,x,y;    while(~scanf("%I64d",&k))    {        int flag=0;        scanf("%I64d%I64d",&a1,&r1);        for(int i=0;i<k-1;i++)        {            scanf("%I64d%I64d",&a2,&r2);            if(flag) continue;            c=r2-r1;            d=exgcd(a1,a2,x,y);            if(c%d)            {                flag=1;            }            else            {                LL q=a2/d;                x=(x*c/d%q+q)%q;//x的解可能是负数，                r1=a1*x+r1;                a1=a1*a2/d;//a1,a2的最小公倍数            }        }        if(flag)            printf("-1\n");        else            printf("%I64d\n",r1);    }    return 0;}