5: Decompose

5: Decompose

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Result	TIME Limit	MEMORY Limit	Run Times	AC Times	JUDGE
	3s	8192K	318	182	Standard

Give you an positive integer N(1<=N<=30), you can decompose n to several positive integers: A1, A2, ... Ak. (1<=k<=N) And A1 + A2 + ... + Ak = N. Now i want to know what's the maximal product of these k integers.

Input

The input contains several test cases. For each test case it contains a positive integer N. The end of input is indicated by end-of-file.

Ouput

For each test case, output K, here K is the maximal product of decomposed numbers.

Sample Input

3

5

6

Sample Output

3

6

9

Problem Source: sharang

 

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This problem is used for contest: 43 

 

 

此类题目，要拆分成小质数。一般情况下就是拆分成多少个3或是多少个2。这是基本的思路。 
假设拆分成的数字里面不只是质数，如里面有6，那么6可以拆成两个3，很显然，乘以6要小于乘以3乘3。 
先从小的数字开始进行分析：4拆分成2+2，2×2=4，这是最大积，4也可以不拆分。5，拆分成3+2得到的乘积是最大的，6拆分成3+3得到的乘积是最大的。分析到这时就可以了，结论就是拆分成3的和，当最后余数是1的时候，那么就拆成最后一个数是4，或者是最后是两个2。当余数是2的时候，就不用特殊变动。 

本题：25÷3=8余1，所以拆分成7个3和1个4（2个2）。

 

 

#include<stdio.h>

#include<math.h>



int n;

double ans;



int main(){



    while(scanf("%d",&n)!=EOF){

        ans=1;

        if(n==1){

            printf("1\n");

            continue;

        }

        if(n%3==1){

            ans=pow(3,(n-4)/3);

            ans*=4;

        }else if(n%3==2){

            ans=pow(3,n/3);

            ans*=2;

        }else

            ans=pow(3,n/3);

        printf("%.0lf\n",ans);

    }

    return 0;

}