E - Power Calculus POJ - 3134

2-Day2-E
Starting with x and repeatedly multiplying by x, we can compute x31 with thirty multiplications:

x2 = x × x, x3 = x2 × x, x4 = x3 × x, …, x31 = x30 × x.

The operation of squaring can be appreciably shorten the sequence of multiplications. The following is a way to compute x31 with eight multiplications:

x2 = x × x, x3 = x2 × x, x6 = x3 × x3, x7 = x6 × x, x14 = x7 × x7, x15 = x14 × x, x30 = x15 × x15, x31 = x30 × x.

This is not the shortest sequence of multiplications to compute x31. There are many ways with only seven multiplications. The following is one of them:

x2 = x × x, x4 = x2 × x2, x8 = x4 × x4, x8 = x4 × x4, x10 = x8 × x2, x20 = x10 × x10, x30 = x20 × x10, x31 = x30 × x.

If division is also available, we can find a even shorter sequence of operations. It is possible to compute x31 with six operations (five multiplications and one division):

x2 = x × x, x4 = x2 × x2, x8 = x4 × x4, x16 = x8 × x8, x32 = x16 × x16, x31 = x32 ÷ x.

This is one of the most efficient ways to compute x31 if a division is as fast as a multiplication.

Your mission is to write a program to find the least number of operations to compute xn by multiplication and division starting with x for the given positive integer n. Products and quotients appearing in the sequence should be x to a positive integer’s power. In others words, x−3, for example, should never appear.

Input
The input is a sequence of one or more lines each containing a single integer n. n is positive and less than or equal to 1000. The end of the input is indicated by a zero.

Output
Your program should print the least total number of multiplications and divisions required to compute xn starting with x for the integer n. The numbers should be written each in a separate line without any superfluous characters such as leading or trailing spaces.

Sample Input
1
31
70
91
473
512
811
953
0
Sample Output
0
6
8
9
11
9
13
12

题意：用最少次数的乘除得到xn。

思路：dfs加剪枝算法，我不太会，讲不明白，学习中。

代码如下：

#include<cstdio>
using namespace std;
const int maxn=1000,maxp=2048;

int n,ans,now[maxn+5];

bool IDAX(int dep)
{
    if (dep>ans) return false;
    if ((now[dep]<<ans-dep)<n) return false; 
    if (now[dep]==n) return true;
    for (int i=0;i<=dep;i++)
    {
        now[dep+1]=now[dep]+now[i];if (IDAX(dep+1)) return true;
        now[dep+1]=now[dep]-now[i];if (IDAX(dep+1)) return true;
    }
    return false;
}
int main()
{
    freopen("program.in","r",stdin);
    freopen("program.out","w",stdout);
    now[0]=1;
    for (scanf("%d",&n);n;scanf("%d",&n))
    {
        for (ans=0;!IDAX(0);ans++);
        printf("%d\n",ans);
    }
    return 0;
}