poj 3134 Power Calculus(迭代加深dfs)

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本文介绍了一种通过使用乘法和除法操作来高效计算x^n的方法，并提供了一个示例程序来展示如何找到从x到x^n所需的最少操作次数。

J - Power Calculus

Time Limit:5000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

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Description

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

Sample Output

题意：从1到n的最小转化步数
思路：迭代加深,深度即步数

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
using namespace std;
const int MAXN=50+5;
int da[]={1,2,4,8,16,32,64,128,256,512,1024};
int p[MAXN],cnt,D,n;
int MAX(const int a,const int b){	return a>b?a:b;}

bool dfs(int d,int sum)
{
	if(d==D)
	{
		if(sum==n)return 1;
		return 0;
	}
	int i,j;
	for(i=cnt-1;i>=0;i--)
	{
		int max=0;
		for(j=0;j<cnt;j++)
			max=MAX(max,p[j]);
		if((max+sum)<<(D-d-1)<n)return 0;//剪枝，减去一定达不到n的搜索

		p[cnt++]=sum+p[i];//加法情况
		if(dfs(d+1,sum+p[i]))return 1;
		cnt--;

		if(sum-p[i]>0)
		{
			p[cnt++]=sum-p[i];//减法情况
			if(dfs(d+1,sum-p[i]))return 1;
			cnt--;
		}
	}
	return 0;
}
int solve(int x)
{
	int i;
	for(i=0;i<=10;i++)
		if(da[i]>=x){D=i;break;}
	while(1)
	{
		memset(p,0,sizeof(p));
		p[0]=1;cnt=1;
		if(dfs(0,1))break;
		D++;
	}
	return D;
}
int main()
{
	while(~scanf("%d",&n),n)
	{
		printf("%d\n",solve(n));
	}
	return 0;
}