Hello Kiki（中国剩余定理——不互质的情况）

最新推荐文章于 2020-10-16 12:08:34 发布

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原文链接：http://www.cnblogs.com/wuwangchuxin0924/p/6327006.html

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#java

本文介绍了一道关于计数问题的编程挑战，名为HelloKiki。该问题要求根据一系列分组后的余数来推算原始数量。通过使用中国剩余定理解决此问题，并提供了完整的代码实现。

Hello Kiki

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 247 Accepted Submission(s): 107

Problem Description

One day I was shopping in the supermarket. There was a cashier counting coins seriously when a little kid running and singing \\\\\\\"门前大桥下游过一群鸭，快来快来数一数，二四六七八\\\\\\\". And then the cashier put the counted coins back morosely and count again...
Hello Kiki is such a lovely girl that she loves doing counting in a different way. For example, when she is counting X coins, she count them N times. Each time she divide the coins into several same sized groups and write down the group size Mi and the number of the remaining coins Ai on her note.
One day Kiki\\\\\\\'s father found her note and he wanted to know how much coins Kiki was counting.

Input

The first line is T indicating the number of test cases.
Each case contains N on the first line, Mi(1 <= i <= N) on the second line, and corresponding Ai(1 <= i <= N) on the third line.
All numbers in the input and output are integers.
1 <= T <= 100, 1 <= N <= 6, 1 <= Mi <= 50, 0 <= Ai < Mi

Output

For each case output the least positive integer X which Kiki was counting in the sample output format. If there is no solution then output -1.

Sample Input

2
2
14 57
5 56
5
19 54 40 24 80
11 2 36 20 76

Sample Output

Case 1: 341
Case 2: 5996

Author

digiter (Special Thanks echo)

Source

2010 ACM-ICPC Multi-University Training Contest（14）——Host by BJTU

Recommend

zhouzeyong

/*
题意：总共有X个钱，分成Mi分会剩余Ai个，让你求X，如果给出的信息不能求出X输出-1

初步思路：中国剩余定理，以前做过类似的题目，韩信点兵，x=m1*m-1*a1+...+mk*mk-1*ak(mod m)
*/
#include<bits/stdc++.h>
#define ll long long 
using namespace std;
/************************中国剩余定理（不互质模板）*****************************/

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 gcd(ll a,ll b)
{
    if(!b){return a;}
    gcd(b,a%b);
}

ll M[55],A[55];


ll China(int r)
{
    ll dm,i,a,b,x,y,d;
    ll c,c1,c2;
    a=M[0];
    c1=A[0];
    for(i=1; i<r; i++)
    {
        b=M[i];
        c2=A[i];
        exgcd(a,b,d,x,y);
        c=c2-c1;
        if(c%d) return -1;//c一定是d的倍数，如果不是，则，肯定无解
        dm=b/d;
        x=((x*(c/d))%dm+dm)%dm;//保证x为最小正数//c/dm是余数，系数扩大余数被
        c1=a*x+c1;
        a=a*dm;
    }
    if(c1==0)//余数为0，说明M[]是等比数列。且余数都为0
    {
        c1=1;
        for(i=0;i<r;i++)
            c1=c1*M[i]/gcd(c1,M[i]);
    }
    return c1;
}

/************************中国剩余定理（不互质模板）*****************************/
int t,n;
int main(){
    //freopen("in.txt","r",stdin);
    scanf("%d",&t);
    for(int ca=1;ca<=t;ca++){
        printf("Case %d: ",ca);
        scanf("%d",&n);
        for(int i=0;i<n;i++) scanf("%lld",&M[i]);
        for(int i=0;i<n;i++) scanf("%lld",&A[i]);
        printf("%lld\n",China(n));
    }
    return 0;
}