POJ_2891_Strange Way to Express Integers(非互质的中国剩余定理)

Strange Way to Express Integers

Time Limit: 1000MS Memory Limit: 131072KTotal Submissions: 8237 Accepted: 2466

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

Hint

All integers in the input and the output are non-negative and can be represented by 64-bit integral types.

Source

POJ Monthly--2006.07.30, Static

题型：数论（非互质的中国剩余定理）

题意：求同余方程组的解

分析：

一眼就看出跟中国剩余定理有关，但是不能用，因为{ a1,a2,……an }不一定是互质的，那肿麽办咧？

采用两两合并的思想：
                                                X ≡ r1 ( mod a1 )
                                                X ≡ r2 ( mod a2 )
推得：
                                                X = k1 * a1 + r1
                                                X = k2 * a2 + r2
所以：                             k1 * a1 + r1 = k2 * a2 + r2
推得：                             k1 * a1 ≡ ( r2 - r1 ) ( mod a2 )
解模线性方程求得最小的 k1
然后得到：                              X = k1 * a1 + r1
令：                                 r = x 、a = lcm ( a1 , a2 )
然后就可以组成另一个式子：
                                                X ≡ r ( mod a )
将这个式子与下一个式子继续像上述一样合并，一直到最后一个并完，r 就是答案。

代码：

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#define LL long longusing namespace std;LL gcd(LL a,LL b){    return b ? gcd(b,a%b) : a;}LL lcm(LL a,LL b){    return a/gcd(a,b)*b;}LL extgcd(LL a,LL b,LL & x,LL & y){    if(b==0){        x=1;        y=0;        return a;    }    LL d=extgcd(b,a%b,x,y);    LL t=x;    x=y;    y=t-a/b*y;    return d;}/*LL modeq(LL a,LL b,LL n){    LL e,i,d,x,y;    d=extgcd(a,n,x,y);    e=(x*(b/d))%n;    return ((e%n+n)%n);}*/long long modeq(long long a,long long b,long long n){long long x,y,d,b1;d=extgcd(a,n,x,y);b1=(n/d)<0 ? -n/d : n/d;x*=b/d;return (x%b1+b1)%b1;}int main(){    LL t,a,r,a1,r1,a2,r2,k1,d;    while(~scanf("%I64d",&t)){        bool flag=1;        scanf("%I64d%I64d",&a1,&r1);        for(int i=1;i<t;i++){            scanf("%I64d%I64d",&a2,&r2);            d=gcd(a1,a2);            if((r2-r1)%d!=0){                flag=0;            }            else{                k1=modeq(a1,(r2-r1),a2);                r1=k1*a1+r1;                a1*=(a2/d);                //a1=lcm(a1,a2);            }        }        if(!flag){            printf("-1\n");        }        else{            printf("%I64d\n",r1);        }    }    return 0;}