fzoj 1320 Ones(动态规划：水题)

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Problem 1320 Ones

Accept: 567 Submit: 1996
Time Limit: 1000 mSec Memory Limit : 32768 KB

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

210

Sample Output

27

Source

chenyan

给出一个数，找到一个仅由‘1’， ‘*'，‘（’，‘）’构成的等式

输出最少需要一个1

很水的一道题，重新开始刷我的DP

状态转移方程为：dp[i] = min(dp[j]+dp[i/j]+dp[i%j])

代码如下：

#include <cstdio>#include <iostream>#include <algorithm>#define MAXN 10010using namespace std;int dp[MAXN];void init(int n) {    for(int i=1; i<=n; ++i) {        dp[i] = i;    }    dp[0] = 0;    for(int i=1; i<=n; ++i) {        for(int j=1; j<=i; ++j) {            dp[i] = min(dp[i%j]+dp[j]+dp[i/j], dp[i]);        }    }}int main(void) {    init(10010);    int n;    while(scanf("%d", &n) != EOF) {        printf("%d\n", dp[n]);    }    return 0;}

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