poj 3134 Power Calculus iddfs(迭代深搜)

iddfs入门题。

//poj 3134//sep9#include <iostream>using namespace std;int n,deep;int a[30];bool iddfs(int pos){int t;if(pos>deep) return false;if(a[pos]<<(deep-pos)<n) return false;if(a[pos]==n)return true;for(int i=1;i<=pos;++i){a[pos+1]=a[i]+a[pos];if(a[pos]+1<=3000&&iddfs(pos+1)==true) return true;a[pos+1]=a[pos]-a[i];if(a[pos]+1>0&&iddfs(pos+1)==true) return true;}return false;}int main(){while(scanf("%d",&n)==1&&n){deep=1,a[1]=1;while(iddfs(1)==0) ++deep;printf("%d\n",deep-1);}return 0;}