HDU 5646 DZY Loves Partition (数学)

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5646

DZY Loves Partition

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)

Total Submission(s): 887    Accepted Submission(s): 327

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

   
   
   
   
   
   
   
   


   
   
   
   

题意：将n个数拆分成k个不同的正整数相加，求这n个正整数的最大乘积

方法：sum（1,k）为k个数相加的最小值，如果n小于这个数，则n无法拆分成k个数， 若n > sum(1, k) 则存在k个连续的数或相差为2的连续上升的数列，  令n = n - sum(1,k)  , c = n / k , d = n % k, 如果d！=0 则存在相差为2数， 所以此时将数列分为两部分，前一部分为连续的数列，后一部分为相差为2的数列，将他们累乘。

#include <bits/stdc++.h>
#define Mod 1000000007
using namespace std;

int main()
{
    int T;
    long long n, k;
    scanf("%d",&T);
    while(T-- && scanf("%lld %lld",&n,&k))
    {
        long long sum = (k + 1) * k / 2;
        if(sum > n)
        {
            printf("-1\n");
            continue;
        }
        n -= sum;
        long long c = n / k;
        long long d = n % k;
        long long ans = 1;
        for(int i=c+1; i<=k+c; i++)
        {
            if(i <= k-d+c)  ans = (ans * i) % Mod;
            else   ans = (ans * (i+1)) % Mod;
        }
        printf("%lld\n", ans%Mod);
    }
    return 0;
}