HDU 5646 DZY Loves Partition(BC)

最新推荐文章于 2019-09-25 17:59:19 发布

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最新推荐文章于 2019-09-25 17:59:19 发布

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本文介绍了一个数学问题，探讨如何将一个正整数n拆分成k个不同的正整数之和，并求出这k个数相乘的最大值。文章提供了具体的解题思路及C语言实现代码。

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DZY Loves Partition

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1414 Accepted Submission(s): 536

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

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 $110888111$.

Source

BestCoder Round #76 (div.2)

分析：

一个正整数n拆分成k个正整数相加，求出k个正整数相乘的最大值。

要取得k个正整数相乘有最大值，即n去和n/k最相近的k个满足要求的数。

先将n/k个整数分配，再分配n%k剩下的数字，即将这些数字补到最大的几个数字上，每个数字加1，如果补到小的数字上会出现与后面重复的情况。

代码如下：

#include <stdio.h>
typedef long long LL;
int a[100005];
int main()
{
	int T;
	int n,k,i,j;
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d %d",&n,&k);
		LL tmp = (LL)k*(k+1)/2;
		if(tmp>n)
		{//如何最小的k个和大于n，则输出-1 
			printf("-1\n");
			continue;
		}
		//将最小的1~k分配到这k个数上 
		for(i=1;i<=k;i++) 
			a[i]=i;
		n-=tmp;		//减去分配的数字
		for(i=1;i<=k;i++)
			a[i]+=n/k;
		n=n%k;
		//来实现将n%k的数补到后面数的几个数上，每个数加一 
		for(i=k;n--;i--)
			a[i]++;	
		LL ans=1; 
		for(i=1;i<=k;i++)
			ans=(ans*a[i])%1000000007;
		printf("%I64d\n",ans);
	}
	
	return 0;
}