uva 1374 - Power Calculus(迭代深搜)

来源：互联网发布：舞美设计用什么软件编辑：程序博客网时间：2024/06/11 02:30

题目链接：uva 1374 - Power Calculus

题目大意：给出n，问说至少计算几步得到x^n。

解题思路：迭代深搜，枚举步数，然后深搜判断是否可行。需要优化，当当前数s按照最大方案执行后仍然小于n，则说明不可行。

#include <stdio.h>#include <string.h>#include <stdlib.h>const int N = 3005;int aid, tmp, v[N], rec[N], pTow[50], ans[N];bool dfs(int d, int s) {if (d == tmp && s == aid) return true;if (d >= tmp || s > 1500) return false;if (s * pTow[tmp - d] < aid) return false;rec[d] = s; v[s] = 1;for (int i = 0; i <= d; i++) {int u = rec[i] + s;if (v[u] == 0)if (dfs(d + 1, u)) return true;u = abs(s - rec[i]);if (v[u] == 0)if (dfs(d + 1, u)) return true;}v[s] = 0;return false;}int solve() {tmp = 0;memset(v, 0, sizeof(v));while (true) {if (dfs(0, 1)) break;tmp++;}return tmp;}void init() {pTow[0] = 1;for (int i = 1; i <= 31; i++) pTow[i] = pTow[i - 1] * 2;}int main() {init();while (scanf("%d", &aid) == 1 && aid) {printf("%d\n", solve());}return 0;}

1 0