暴力求解——POJ 3134Power Calculus

Description

Starting with x and repeatedly multiplying by x, we can compute x³¹ with thirty multiplications:

x² = xxx,     x³ = x²xx,     x⁴ = x³xx,     ...  ,     x³¹ = x³⁰xx.

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

x² = xxx,     x³ = x²xx,     x⁶ = x³xx³,     x⁷ = x⁶xx,     x¹⁴ = x⁷xx⁷,
x¹⁵ = x¹⁴xx,     x³⁰ = x¹⁵xx¹⁵,     x³¹ = x³⁰xx.

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 = xxx,     x⁴ = x²xx²,     x⁸ = x⁴xx⁴,     x¹⁰ = x⁸xx²,
x²⁰ = x¹⁰xx¹⁰,     x³⁰ = x²⁰xx¹⁰,     x³¹ = x³⁰xx.

There however is no way to compute  x³¹ with fewer multiplications. Thus this is one of the most efficient ways to compute  x³¹ only by multiplications.

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

x² = xxx,     x⁴ = x²xx²,     x⁸ = x⁴xx⁴,     x¹⁶ = x⁸xx⁸,     x³² = x¹⁶xx¹⁶,     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 of operations should be x to a positive integer's power. In other 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

解题思路：

题目大意：求用一个x，如何在最小的步数使用已经用过的凑出x^n，其中n是1~1000的，可以搜索，很容易想到，<=0和>=2000的点应该直接剪掉。

这个题目方根就是搜索的时候控制搜索的层数即可。

 if(a[c]*(1<<(t-c))<n)
        return false;    //这个约束条件必须加，剪枝，这个点按最大比例每次都乘以2扩充也扩充不到n，那么就剪掉。

程序代码：

#include <iostream>
using namespace std;
int a[1010];
int t,b,n;
bool funt(int c)
{
    if(c>t)
        return false;
    if(a[c]==n)
        return true;
    if(a[c]<=0||a[c]>10000)
        return false;
    if(a[c]*(1<<(t-c))<n)
        return false;
    for(int i=0;i<=c;i++)
    {
        a[c+1]=a[c]+a[i];
        if(funt(c+1))
            return true;
        a[c+1]=a[c]-a[i];
        if(funt(c+1))
            return true;
    }
    return 0;
}

int main()
{
    while(cin>>n&&n)
    {
        if(n==1)
            cout<<0<<endl;
        else
        {
            t=1;
            while(1)
            {
                a[0]=1;
                if(funt(0))
                    break;
                t++;
            }
            cout<<t<<endl;
        }
    }
    return 0;
    
}