poj 3134 Power Calculus （IDA*）

Power Calculus

Time Limit: 5000MS Memory Limit: 65536KTotal Submissions: 1593 Accepted: 835

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

13170914735128119530

Sample Output

06891191312

Source

Japan 2006

题意：

通过乘法和除法将 X 变为 X^n ，找出最快需要几步，每一步相乘或相除的数都需要再次之前出现过。

思路：

IDA*，每次限制深度往下搜，A*剪枝为 设当前的数为x^a，那么这个数通过平方迭代增长最大次数都不能达到n的话就剪枝。增长过程为 x^a->x^2a->x^4a->x^8a...

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <string>#include <map>#include <stack>#include <vector>#include <set>#include <queue>#pragma comment (linker,"/STACK:102400000,102400000")#define maxn 2005#define MAXN 100005#define mod 1000000007#define INF 0x3f3f3f3f#define pi acos(-1.0)#define eps 1e-6typedef long long ll;using namespace std;int n,m,ans,k,flag,cnt,deep;int vis[maxn];void dfs(int pos){    int i,j,t;    if(flag||pos>deep||(vis[pos]<<(deep-pos))<n) return ; // 通过自身相乘(即指数相加 最快方式)都不能达到n 剪枝    if(vis[pos]==n)    {        flag=1;        return ;    }    for(i=1;i<=pos;i++)    {        if(flag) return ;        t=vis[pos]+vis[i];        if(t>0&&t<2000)        {            vis[pos+1]=t;            dfs(pos+1);        }        t=vis[pos]-vis[i];        if(t>0&&t<2000)        {            vis[pos+1]=t;            dfs(pos+1);        }    }}int main(){    int i,j;    while(scanf("%d",&n),n)    {        deep=flag=0;   // deep-多少个数相乘        vis[1]=1;        while(!flag)   // 迭代加深        {            deep++;            dfs(1);        }        printf("%d\n",deep-1);    }    return 0;}