POJ 1320 Ones（普通的DP）

Problem Description

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.

 Input

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

 Output

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

 Sample Input

2

10

 Sample Output

2

7

题意：给你一个数n，要你求出这个数最少由给定的格式1组成！

思路：直接DP推方程了！dp[i]=min(dp[i],dp[j]+[i/j]+dp[i%j])

所以AC代码：

#include<cstdio>#include<cmath>#include<algorithm>using namespace std;const int maxn=10000+10;int dp[maxn];void init(){    for(int i=1;i<maxn;i++)    {        dp[i]=i;        for(int j=1;j*j<=i;j++)        {            dp[i]=min(dp[i],dp[j]+dp[i/j]+dp[i%j]);        }    }}int main(){    int n;    init();    while(scanf("%d",&n)!=EOF)    {        printf("%d\n",dp[n]);    }    return 0;}