NYOJ 708 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

#include<cstdio>
#include<cmath>
#include<iostream>
using namespace std;
#define INF 100003
int dp[10002];
int main()
{
    int i, z, n;
    dp[1] = 1, dp[2] = 2;
    for(i = 3; i < 10001; i++){
        dp[i] = INF;
        for(z = 1; z < i && z <= i / 2; z++){
            dp[i] = min(dp[i], dp[i-z] + dp[z]);
            if(z != 1 && (i%z) == 0){
                dp[i] = min(dp[i], dp[i/z]+dp[z]);
            }
        }
    }
    while(~scanf("%d", &n))
    {
        printf("%d
", dp[n]);
    }
    return 0;
}