nyoj_ones

ones

時間限制：1000 ms  |  內存限制：65535 KB

難度：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.

輸入

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

輸出

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

樣例輸入

2
10

樣例輸出

2
7

       f(n - 1) + f(1);

f(n) = f (n / 2) + f(2);

       f (n / 3) + f(3);

       ...

       除數是100以內的素數

#include <iostream>
#include <cstdio>
#include <cstring>
#define N 10010
using namespace std;

int n;
int res[N];
bool visit[N];
int prime[110];
int len;

void init()
{
    for (int i = 2; i * i < N; i++)
    {
        if (!visit[i])
        {
            prime[len++] = i;
            for (int j = i + i; j < N; j += i)
                visit[j] = 1;
        }
    }
}

int dp(int num)
{
    if (res[num] != 0)
        return res[num];
    if (num == 1)
    {
        res[num] = 1;
        return res[num];
    }
    int a = dp(num - 1) + 1;
    for (int i = 0; prime[i] < num && i < len; i++)
    {
        if (num % prime[i] == 0)
        {
            int temp = dp(num / prime[i]) +dp(prime[i]);
            if (a > temp)
                a = temp;
        }
    }
    res[num] = a;
    return res[num];
}



int main()
{
    init();
    while (scanf("%d", &n) != EOF)
    {
        printf("%d\n", dp(n));
    }

    return 0;
}