HDU 5646

DZY Loves Partition

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 815    Accepted Submission(s): 298

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

2

4

110888111

Hint

In 1st testcase, there is no valid partition. In 2nd testcase, the partition is $3=1+2$. Answer is $1 imes 2 = 2$. In 3rd testcase, the partition is $9=2+3+4$. Answer is $2 imes 3 imes 4 = 24$. 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 $333332 imes 333334= 111110888888$. Remember to output it mudulo $10^9 + 7$, which is $110888111$.

 

Source

BestCoder Round #76 (div.2)

 

题意：t组数据 将n分为k个数的和 使得 这k个数的乘积最大（k个数不能重复） 输出这个数

官方题解

记sum(a,k)=a+(a+1)+⋯+(a+k−1)

首先，有解的充要条件是sum(1,k)≤n（如果没取到等号的话把最后一个k扩大就能得到合法解）。

然后观察最优解的性质，它一定是一段连续数字，或者两段连续数字中间只间隔1个数。这是因为1≤a<=b−2时有ab<(a+1)(b−1)如果没有满足上述条件的话，我们总可以把最左边那段的最右一个数字作为a，最右边那段的最左一个数字作为b，调整使得乘积更大。

可以发现这个条件能够唯一确定n的划分，只要用除法算出唯一的a使得sum(a,k)≤n<sum(a+1,k)就可以得到首项了。

理解我yan代码  先排列一个k位的首项为1 的等差数列 将剩下的数 优先靠右分配

       

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<map>
using namespace std;
#define mod 1000000007
#define  ll __int64
ll T,n,k,f;
ll ans = 1;
int main() {
    scanf("%I64d",&T);
    while(T--) 
    {
        scanf("%I64d%I64d",&n,&k);
        ans = 1;
        if(k==1) 
        {
            cout<<n<<endl;
            continue;
        }
        if(n<k) 
        cout<<-1<<endl;
        else 
        {
            n = n-(1+k)*k/2;//每个只能出现一次 
            if(n<0) 
        {
                cout<<-1<<endl;continue;
        }
            if(n==0) 
            {
                for(int i=1;i<=k;i++) 
                ans=(ans*i)%mod;
                cout<<ans<<endl;
                continue;
            }
           for(int i=k;i>k-n%k;i--)
               ans=(ans*(n/k+1+i))%mod;
           for(int i=k-n%k;i>=1;i--) 
                ans=(ans*(n/k+i))%mod;
           cout<<ans<<endl;
        }
    }
}