HDU 3579 Hello Kiki 解题报告（中国剩余定理 非互质）

Hello Kiki

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1835    Accepted Submission(s): 649

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)

 

    解题报告：在刘汝佳老师的书上学了中国剩余定理，不过限制是给定的m[]都是两两互质的。本题中的除数就不是互质的，所以我们要换一种方法：合并模方程。

    关于合并的证明，我在这里看到的： http://yzmduncan.iteye.com/blog/1323599/。建议大家自己证明一下。

    另外，题目有个坑爹的地方，输出正数。如果结果是0，就加上最后的模数。

    本人代码如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int maxn = 100;
int mi[maxn];
int ai[maxn];
int cas=1;

void gcd(int a,int b,int &d,int &x,int &y)
{
    if(b==0)
    {
        d=a;
        x=1;
        y=0;
    }
    else
    {
        gcd(b,a%b,d,y,x);
        y-=x*(a/b);
    }
}

int china_2(int n,int m[],int a[])
{
    int aa = a[0];
    int mm = m[0];

    for(int i=1;i<n;i++)
    {
        int ta=((a[i]-aa)%m[i]+m[i])%m[i];

        int d,x,y;
        gcd(mm,m[i],d,x,y);
        if(ta%d) return -1;

        int newM = mm/d*m[i];
        aa = ((long long)ta/d*x*mm%newM+aa)%newM;
        mm = newM;
    }
    aa = (aa+mm)%mm;
    if(aa==0) aa=mm;
    return aa;
}

void work()
{
    int n;
    scanf("%d",&n);

    for(int i=0;i<n;i++)
        scanf("%d", mi+i);
    for(int i=0;i<n;i++)
        scanf("%d", ai+i);

    int ans = china_2(n,mi,ai);
    printf("Case %d: %d\n", cas++, ans);
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
        work();
}