NYOJ 708 ones(基础dp)(try again)

ones

时间限制： 1000 ms  |  内存限制： 65535 KB

难度： 3

描述

Given a positive integer N (0<=N<=10000), you are to find an expression equals to N using only 1,+,*,(,). 1 should not appear continuously, i.e. 11+1 is not allowed.

输入

There are multiple test cases. Each case contains only one line containing a integer N

输出

For each case, output the minimal number of 1s you need to get N.

样例输入

2
10

样例输出

2
7

//最优子结构就是，21调用3*7的时候7已经是最优状态，然后直接调用

//不是完全理解

#include
#include
#include
#include

using namespace std;
int dp[10010]={0};
int main()
{
	dp[1]=1;
	for(int i=2; i<10010; i++)
	{
		dp[i]=dp[i-1]+1;
		for(int j=i-1; j>=1; j--)
		{
			if(i%j==0)
			{
				dp[i]=min(dp[i],dp[j]+dp[i/j]);
			}
		}
	}
	int n;
	while(~scanf("%d",&n))
	{
		printf("%d\n",dp[n]);
	}
	return 0;
} //ac