UVA-1374 Power Calculus (迭代加深搜索)
题目大意:问最少经过几次乘除法可以使x变成xn。
题目分析:迭代加深搜索。
代码如下:
# include<iostream>
# include<cstdio>
# include<set>
# include<cstring>
# include<algorithm>
using namespace std;
int num[20],maxn,cnt;
bool dfs(int d,int maxd,int n)
{
if(d==maxd)
return num[cnt-1]==n;
if((num[cnt-1]<<(maxd-d))<n)
return false;
for(int i=0;i<cnt;++i){
num[cnt]=num[cnt-1]+num[i];
++cnt;
if(num[cnt-1]<=1000&&dfs(d+1,maxd,n))
return true;
--cnt;
num[cnt]=num[cnt-1]-num[i];
++cnt;
if(num[cnt-1]>0&&dfs(d+1,maxd,n))
return true;
--cnt;
}
return false;
}
int main()
{
int n;
while(scanf("%d",&n)&&n)
{
for(int maxd=0;;++maxd){
cnt=0;
num[cnt++]=1;
if(dfs(0,maxd,n)){
printf("%d\n",maxd);
break;
}
}
}
return 0;
}
打表版本:
# include<iostream>
# include<cstdio>
# include<set>
# include<cstring>
# include<algorithm>
using namespace std;
int ans[1000]={0,1,2,2,3,3,4,3,4,4,5,4,5,5,5,4,5,5,6,5,6,6,6,5,6,6,6,6,7,6,6,5,6,6,
7,6,7,7,7,6,7,7,7,7,7,7,7,6,7,7,7,7,8,7,8,7,8,8,8,7,8,7,7,6,7,7,8,7,8,8,8,7,8,8,8,8,
8,8,8,7,8,8,8,8,8,8,9,8,9,8,9,8,8,8,8,7,8,8,8,8,9,8,9,8,9,9,9,8,9,9,9,8,9,9,9,9,9,9,9,8,
9,9,9,8,9,8,8,7,8,8,9,8,9,9,9,8,9,9,9,9,9,9,9,8,9,9,9,9,9,9,10,9,9,9,9,9,9,9,9,
8,9,9,9,9,9,9,10,9,10,9,10,9,10,10,10,9,10,10,10,9,10,10,10,9,10,9,10,9,9,9,9,8,
9,9,9,9,10,9,10,9,10,10,10,9,10,10,10,9,10,10,10,10,10,10,10,9,10,10,10,10,10,10,
10,9,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,9,10,10,10,10,10,10,10,9,10,10,
10,9,10,9,9,8,9,9,10,9,10,10,10,9,10,10,11,10,11,10,10,9,10,10,11,10,11,10,10,10,
10,10,10,10,10,10,10,9,10,10,10,10,10,10,11,10,10,10,11,10,11,11,11,10,11,10,11,10,
11,10,11,10,11,10,10,10,10,10,10,9,10,10,10,10,10,10,11,10,11,10,11,10,11,11,11,10,11,11,11,
10,11,11,11,10,11,11,11,11,11,11,11,10,11,11,11,11,11,11,11,10,11,11,11,11,11,11,11,
10,11,11,11,10,11,11,11,10,11,10,11,10,10,10,10,9,10,10,10,10,11,10,11,10,11,11,11,10,11,
11,11,10,11,11,11,11,11,11,11,10,11,11,11,11,11,11,11,10,11,11,11,11,11,11,11,11,
11,11,11,11,11,11,11,10,11,11,11,11,11,11,11,11,11,11,11,11,11,11,11,10,11,11,11,11,11,
11,11,11,11,11,11,11,12,11,12,11,11,11,12,11,12,11,11,11,11,11,11,11,11,11,11,10,11,11,
11,11,11,11,11,11,11,11,12,11,12,11,11,10,11,11,12,11,12,11,11,10,11,11,11,10,11,10,10,9,10,
10,11,10,11,11,11,10,11,11,12,11,12,11,11,10,11,11,12,11,12,12,11,11,12,12,12,11,12,11,11,10,
11,11,12,11,12,12,12,11,11,12,12,11,12,11,12,11,11,11,12,11,12,11,11,11,12,11,11,11,11,11,11,10,11,
11,11,11,11,11,12,11,11,11,12,11,12,12,12,11,12,11,12,11,12,12,12,11,12,12,12,12,12,
12,12,11,12,12,12,11,12,12,12,11,12,12,12,11,12,12,12,11,12,12,12,11,12,11,12,11,
12,11,11,11,11,11,11,10,11,11,11,11,11,11,12,11,12,11,12,11,12,12,
12,11,12,12,12,11,12,12,12,11,12,12,12,12,12,12,12,11,12,12,12,12,12,12,12,11,12,12,
12,12,12,12,12,11,12,12,12,12,12,12,12,12,12,12,12,12,12,12,12,11,12,12,12,12,12,12,
12,12,12,12,12,12,12,12,12,11,12,12,12,12,12,12,12,12,12,12,12,12,12,12,
12,11,12,12,12,12,12,12,12,11,12,12,12,12,12,12,12,11,12,12,12,11,12,12,12,11,12,
11,12,11,11,11,11,10,11,11,11,11,12,11,12,11,12,12,12,11,12,12,12,11,12,12,12,12,
12,12,12,11,12,12,12,12,12,12,12,11,12,12,12,12,12,12,12,12,12,12,13,12,12,12,
12,11,12,12,12,12,13,12,12,12,12,12,12,12,12,12,12,11,12,12,12,12,12,12,12,12,12,
12,12,12,12,12,13,12,12,12,13,12,13,12,12,12,13,12,12,12,12,12,12,11,12,12,12,12,12,
12,13,12,12,12,13,12,13,12,12,12,13,12,13,12,13,12,12,12,12,12,12,12,12,12,
12,11,12,12,12,12,12,12,12,12,12,12,13,12,13,12,13,12,13,12,13,12,13,12,13,12,13,
13,13,12,13,13,13,12,13,12,13,12,13,13,13,12,13,13,13,12,13,12,13,12,12,12,13,12,
13,12,12,12,12,12,12,12,12,12,12,11,12,12,12,12,12,12,12,12,12,12,13,12,13,12,
12,12,12,12,13,12,13,13,13,12,13,13,13,12,13,12,12,11,12,12,13,12,13,13,13,12};
int main()
{
int n;
while(scanf("%d",&n)&&n)
printf("%d\n",ans[n-1]);
return 0;
}
UVA-1374 Power Calculus (迭代加深搜索)的更多相关文章
- UVA - 1374 Power Calculus (dfs迭代加深搜索)
题目: 输入正整数n(1≤n≤1000),问最少需要几次乘除法可以从x得到xn ?在计算过程中x的指数应当总是正整数. 思路: dfs枚举次数深搜 注意: 1.指数如果小于0,就退出当前的搜索 2.n ...
- UVa 1374 - Power Calculus——[迭代加深搜索、快速幂]
解题思路: 这是一道以快速幂计算为原理的题,实际上也属于求最短路径的题目类型.那么我们可以以当前求出的幂的集合为状态,采用IDA*方法即可求解.问题的关键在于如何剪枝效率更高.笔者采用的剪枝方法是: ...
- UVA1374-Power Calculus(迭代加深搜索)
Problem UVA1374-Power Calculus Accept:107 Submit:584 Time Limit: 3000 mSec Problem Description In ...
- POJ 3134 Power Calculus (迭代剪枝搜索)
题目大意:略 题目里所有的运算都是幂运算,所以转化成指数的加减 由于搜索层数不会超过$2*log$层,所以用一个栈存储哪些数已经被组合出来了,不必暴力枚举哪些数已经被搜出来了 然后跑$iddfs$就行 ...
- UVA 1374 Power Calculus
题意: 给出m,问对n最少进行几次操作.n初始为1,能得到m.操作1位将n平方.操作2为将n除以之前出现的n值中的任意一个. 分析: 其实是关于指数的操作,即从1到m最少的步数.我们可以先确定最少步数 ...
- UVa 1374 Power Calculus (IDA*或都打表)
题意:给定一个数n,让你求从1至少要做多少次乘除才可以从 x 得到 xn. 析:首先这个是幂级的,次数不会很多,所以可以考虑IDA*算法,这个算法并不难,难在找乐观函数h(x), 这个题乐观函数可以是 ...
- Power Calculus UVA - 1374 迭代加深搜索
迭代加深搜索经典题目,好久不做迭代加深搜索题目,拿来复习了,我们直接对当前深度进行搜索,注意剪枝,还有数组要适当开大,因为2^maxd可能很大 题目:题目链接 AC代码: #include <i ...
- 【算法•日更•第三十九期】迭代加深搜索:洛谷SP7579 YOKOF - Power Calculus 题解
废话不多说,直接上题: SP7579 YOKOF - Power Calculus 题意翻译 (略过没有营养的题干) 题目大意: 给出正整数n,若只能使用乘法或除法,输出使x经过运算(自己乘或除自己, ...
- UVA 529 - Addition Chains,迭代加深搜索+剪枝
Description An addition chain for n is an integer sequence with the following four properties: a0 = ...
随机推荐
- web前端----JavaScript的BOM
一.引入 到目前为止,我们已经学过了JavaScript的一些简单的语法.但是这些简单的语法,并没有和浏览器有任何交互. 也就是我们还不能制作一些我们经常看到的网页的一些交互,我们需要继续学习BOM和 ...
- 计算概论(A)/基础编程练习1(8题)/8:与7无关的数
#include<stdio.h> int main() { ; // n < 100 scanf("%d", &n); // 循环遍历判断 再进行平方和 ...
- 安装使用composer基本流程
composer工作原理: 这里经过几个步骤:1.composer读取composer.json(这个文件手动建立,官网有格式),这个json是在当前执行composer目录的,如果目录下没有这个js ...
- 20145331魏澍琛 《网络对抗技术》 PC平台逆向破解
20145331魏澍琛 <网络对抗技术> PC平台逆向破解 学习任务 1.shellcode注入:shellcode实际是一段代码,但却作为数据发送给受攻击服务器,将代码存储到对方的堆栈中 ...
- 计算TCP链接的RTO超时重传时间
- Spring DBCP用xml和properties2种格式配置DataSource
Spring提供数据库连接池:DBCP配置DataSource并且获取连接完成数据库操作: Spring帮助文档的地址: http://static.springsource.org/spring/d ...
- IT人士级别的划分
IT领袖:年入过亿(例如任正非.马化腾.李彦宏.丁磊.马云等,包括期权股票以及投资理财等收入.) IT大哥:年入千万(级别次于以上几位大佬的公司老板,不缺钱,普遍对上一条里的人物羡慕嫉妒恨.) IT精 ...
- Git基础 —— Github 的使用
Git 基础学习系列 Git 基础 -- 安装 配置 别名 对象 Git 基础 -- 常用命令 Git 基础 -- 常见使用场景 Git基础 -- Github 的使用 Github 的利用 Gith ...
- C#中配置文件保存的路径
http://www.codeproject.com/Tips/350010/Saving-User-Settings-in-Winform-Application 外网上找的资料 winform提供 ...
- ajax中的contentType使用
本文为博主原创,未经允许不得转载: 最近在修改部分项目功能的时候,遇到一个问题.局部刷新某页面的功能是由ajax实现的,但当我进行局部刷新的时候,页面并没有刷新和响应, 在后台的代码中打了断点也并没有 ...