POJ 3134 - Power Calculus (IDDFS)

题意：求仅仅用乘法和除法最快多少步能够求到x^n

思路：迭代加深搜索

//Accepted	164K	1094MS	C++	840B
include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
int step[100005];
int n;
int cur;
bool IDDFS(int lim,int g)
{

    if(cur>lim) return false;
    if(step[cur]==g) return true;
    if((step[cur]<<(lim-cur))<g) return false; //剪枝
    for(int i=0;i<=cur;i++)
    {
        cur++;
        step[cur]=step[i]+step[cur-1];
        if(step[cur]<=10000&&IDDFS(lim,g)) return true; //比1000^2大的时候就没意义了，剪枝
        step[cur]=step[cur-1]-step[i];
        if(step[cur]>0&&IDDFS(lim,g)) return true;
        cur--;
    }
    return false;
}
int main()
{
    while(scanf("%d",&n),n)
    {
        int ans=-1;
        while(1)
        {
            cur=0;
            step[cur] = 1;
            ans++;
            if(IDDFS(ans,n)) break;
        }
        printf("%d\n",ans);
    }
    return 0;
}