Hello Kiki hdu 3579 求解一元线性同余方程组ps:中国剩余定理 Hello Kiki

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概述

Hello Kiki

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1833 Accepted Submission(s): 648

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

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zhouzeyong | We have carefully selected several similar problems for you: 3573 3574 3575 3576 3577

题意是说有个小孩有钱，然后她每次都点m次，给出分的组数和剩余的钱数，求解最小的钱数。这个题目很明显同余方程组的求解，题目要求最小正整数解，如果求出同余方程组的解为0，则输出mi的最小公倍数

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define N  15

long long gcd(long long a,long long b)
{
    return b==0?a:gcd(b,a%b);
}
void exgcd(long long a,long long b,long long &d,long long &x,long long &y)
{
    if(!b)
    {
        x=1;
        y=0;
        d=a;
    }
    else
    {
        exgcd(b,a%b,d,y,x);
        y=y-x*(a/b);
    }
}

long long aa[N],rr[N];
int main()
{
    long long n,a,b,c,d,x,y,lcm;
    
    int T,i,m;
    int e=1;
    scanf("%d",&T);
     for(e=1;e<=T;e++)
    {
        lcm=1;
        scanf("%d",&m);
        int ha=1;
        //printf("aaaaaa");
        //printf("%dn",m);
        for(i=1; i<=m; i++)
        {
            cin>>aa[i];
            //printf("%d %dn",i,m);
            // if(i==m)break;
            //printf("fdsfdsbbbbbbb");
            lcm=lcm/gcd(aa[i],lcm)*aa[i];
        }
       // printf("eeeeeeeeeeeeee");
        for(i=1; i<=m; i++)
        {
            scanf("%lld",&rr[i]);
        }

        printf("Case %d: ",e);
        for(i=2; i<=m; i++)
        {
            a=aa[1];
            b=aa[i];
            c=rr[i]-rr[1];
            exgcd(a,b,d,x,y);
            if(c%d!=0)
            {
                ha=0;
                break;
            }
            long long t=b/d;
            x=(x*(c/d)%t+t)%t;
            rr[1]=aa[1]*x+rr[1];
            aa[1]=aa[1]*(aa[i]/d);
        }
        if(!ha)
        {
            printf("-1n");
            continue;
        }
        if(rr[1]!=0)
        {
            printf("%lldn",rr[1]);
            continue;
        }
        printf("%lldn",lcm);
    }
    return 0;
}