題目鏈接:Power Calculus
題目大意:給出n,求最少多少次乘除操作使
解題思路:由於不知道最後的深度是多少,直接上DFS一定會超時,所以改用迭代加深搜索,在搜索過程中注意剪枝。
代碼如下:
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <cstdio>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef pair<int, int> P;
const int inf = 0x3f3f3f3f;
const int maxn = 2e4 + 15;
int n, dp[1005], cnt, maxd;
bool DFS(int x, int dep){
if(dep >= maxd) return x == n;
int ma = *max_element(dp, dp + cnt);
if(ma << maxd - dep < n) return false;
for(int i = cnt - 1; i >= 0; i--){
dp[cnt++] = x + dp[i];
if(DFS(x + dp[i], dep + 1)) return true;
cnt--;
dp[cnt++] = x - dp[i];
if(DFS(x - dp[i], dep + 1)) return true;
cnt--;
}
return false;
}
int main(){
#ifdef LOCAL
freopen("in.txt", "r", stdin);
#endif
ios::sync_with_stdio(false);
cin.tie(0);
memset(dp, 0, sizeof dp);
while(cin >> n && n){
if(dp[n]) goto Answer;
for(maxd = __lg(n); ; maxd++){
cnt = 1; dp[0] = 1;
if(DFS(1, 0)) break;
}
Answer:
cout << maxd << endl;
}
return 0;
}