【POJ 2891】Strange Way to Express Integers 【 CRT 证明详解】

最新推荐文章于 2019-10-06 13:19:31 发布

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最新推荐文章于 2019-10-06 13:19:31 发布

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分类专栏：结构的模版 gcd + lcm + exgcd + CRT

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gcd + lcm + exgcd + CRT

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本文介绍了解决一种特殊的中国剩余定理(CRT)问题的算法实现，该问题的特点是模数之间并非互质。文章详细阐述了通过合并方程的方法来解决此类问题的过程，并给出了完整的代码实现。

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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
2
8 7
11 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

CRT 问题，不过模之间不是互质的。
这里采取的是合并方程的方式：百度了好多好证明才看懂，为了加强概念，手写证明了好几遍，有点丑TAT 。
证明原文
这里写图片描述
之后就是模板化就好了
看代码

#include<stdio.h>
#define LL long long 
const int MAXN = 1e5;
const int MAXM = 1e6;
const int mod = 1e9+7;
const int inf = 0x3f3f3f3f;

struct CRT{
    LL gcd(LL a,LL b) { return b==0?a:gcd(b,a%b);}
    void exgcd(LL a,LL b,LL &d,LL &x,LL &y){
        if(!b){ d=a; x=1;y=0;}
        else {
            exgcd(b,a%b,d,y,x);
            y-=x*(a/b);
        }
    }
    LL China(LL le,LL ri,LL *m,LL *a){
        LL lcm=1;
        for(LL i=le;i<=ri;i++) 
            lcm=lcm/gcd(lcm,m[i])*m[i];
        for(LL i=le+1;i<=ri;i++){
            LL A=m[le],B=m[i],d,x,y,c=a[i]-a[le];
            exgcd(A,B,d,x,y);
            if(c%d) return -1;
            LL mod=m[i]/d;
            LL k=((x*c/d)%mod+mod)%mod;
            a[le]=m[le]*k+a[le];
            m[le]=m[i]*m[le]/d;
        }
        if(a[le]==0) return lcm;
        return a[le];
    }
}crt;
LL a[MAXN],m[MAXN]; 
int main(){

    LL n;
    while(scanf("%lld",&n)!=EOF){
        for(int i=1;i<=n;i++)
            scanf("%lld%lld",&m[i],&a[i]);
        printf("%lld\n",crt.China(1,n,m,a));
    }
    return 0;
}