HDOJ 5646 DZY Loves Partition

DZY Loves Partition

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 117    Accepted Submission(s): 38

Problem Description

DZY loves partitioning numbers. He wants to know whether it is possible to partition n into the sum of exactly k distinct positive integers.

After some thinking he finds this problem is Too Simple. So he decides to maximize the product of these k numbers. Can you help him?

The answer may be large. Please output it modulo 109+7 .

 

Input

First line contains t denoting the number of testcases.

t testcases follow. Each testcase contains two positive integers n,k in a line.

( 1≤t≤50,2≤n,k≤109 )

 

Output

For each testcase, if such partition does not exist, please output −1 . Otherwise output the maximum product mudulo 109+7 .

 

Sample Input

  
  

   
   4
3 4
3 2
9 3
666666 2
  
  

 

Sample Output

  
  

   
   -1
2
24
110888111


   
   

    
    

     
     Hint
    
    
In 1st testcase, there is no valid partition.
In 2nd testcase, the partition is $3=1+2$. Answer is $1\times 2 = 2$.
In 3rd testcase, the partition is $9=2+3+4$. Answer is $2\times 3 \times 4 = 24$. Note that $9=3+3+3$ is not a valid partition, because it has repetition.
In 4th testcase, the partition is $666666=333332+333334$. Answer is $333332\times 333334= 111110888888$. Remember to output it mudulo $10^9 + 7$, which is $110888111$.

   
   
   
    
  
  

#include <iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#define MOD 1000000007
using namespace std;

int main()
{
    int t,i;
    __int64 s,n,k;
    scanf("%d",&t);
    while(t--)
    {
        s=1;
        scanf("%I64d%I64d",&n,&k);
        if(n<k*(k+1)/2)
        {
            printf("-1\n");
            continue;
        }
        if(n%k==0)
        {
            int tem=n/k;
            if(k&1)
                s=tem;//这里竟然写成k。。。。我真是个。。。。
            for(i=1;i<=k/2;++i)
                s=s*(tem-i)*(tem+i)%MOD;
        }
        else
        {
            int tem=k*(k-1)/2;//这里将第一个数看做基数，这里加和的是之后的k-1个数在基数上的增量（这里先将这k个数看为连续的）
            int num=(n-tem)%k;//将这些增量都减去再取余就得到若是这k个数如此取加和之后与n的差量（需将这些差量补充到k个数中）
            int base=(n-tem)/k;//求基数。因为要求乘积最大所以这些增量需要一个一个的加到后num个数上）
            i=1;
            for(;i<=num;++i)
            {
                s=s*(base+k-i+1)%MOD;//这里就是要加1的乘积
            }
            while(i<=k)
                s=s*(base+k-i++)%MOD;//不加1的乘积
        }
        printf("%I64d\n",s);
    }
    return 0;
}