Safecracker

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12702    Accepted Submission(s): 6581

Problem Description
=== Op tech briefing, 2002/11/02 06:42 CST === 
"The item is locked in a Klein safe behind a painting in the second-floor library. Klein safes are extremely rare; most of them, along with Klein and his factory, were destroyed in World War II. Fortunately old Brumbaugh from research knew Klein's secrets and wrote them down before he died. A Klein safe has two distinguishing features: a combination lock that uses letters instead of numbers, and an engraved quotation on the door. A Klein quotation always contains between five and twelve distinct uppercase letters, usually at the beginning of sentences, and mentions one or more numbers. Five of the uppercase letters form the combination that opens the safe. By combining the digits from all the numbers in the appropriate way you get a numeric target. (The details of constructing the target number are classified.) To find the combination you must select five letters v, w, x, y, and z that satisfy the following equation, where each letter is replaced by its ordinal position in the alphabet (A=1, B=2, ..., Z=26). The combination is then vwxyz. If there is more than one solution then the combination is the one that is lexicographically greatest, i.e., the one that would appear last in a dictionary."

v - w^2 + x^3 - y^4 + z^5 = target

"For example, given target 1 and letter set ABCDEFGHIJKL, one possible solution is FIECB, since 6 - 9^2 + 5^3 - 3^4 + 2^5 = 1. There are actually several solutions in this case, and the combination turns out to be LKEBA. Klein thought it was safe to encode the combination within the engraving, because it could take months of effort to try all the possibilities even if you knew the secret. But of course computers didn't exist then."

=== Op tech directive, computer division, 2002/11/02 12:30 CST ===

"Develop a program to find Klein combinations in preparation for field deployment. Use standard test methodology as per departmental regulations. Input consists of one or more lines containing a positive integer target less than twelve million, a space, then at least five and at most twelve distinct uppercase letters. The last line will contain a target of zero and the letters END; this signals the end of the input. For each line output the Klein combination, break ties with lexicographic order, or 'no solution' if there is no correct combination. Use the exact format shown below."

 
Sample Input
1 ABCDEFGHIJKL
11700519 ZAYEXIWOVU
3072997 SOUGHT
1234567 THEQUICKFROG
0 END
 
Sample Output
LKEBA
YOXUZ
GHOST
no solution

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int MAXN=;
int target,len;
char s[MAXN];
int vis[MAXN];
int buf[MAXN];
bool comp(char ch1,char ch2)
{
return ch1 > ch2;
}
bool dfs(int dep)
{
if(dep==)
{
long long sum=;
for(int i=;i<dep;i++)
{
long long e=;
for(int j=;j<=i;j++)
{
e*=buf[i];
}
if(i%==) sum+=e;
else sum-=e;
}
if(sum==(long long)target)
{
return true;
}
else
{
return false;
}
}
for(int i=;i<len;i++)
{
if(!vis[i])
{
vis[i]=;
buf[dep]=s[i]-'A'+;
if(dfs(dep+))
{
return true;
}
vis[i]=;
}
}
return false;
}
int main()
{
while(scanf("%d%s",&target,s)!=EOF)
{
if(target==&&strcmp(s,"END")==) break;
memset(vis,,sizeof(vis));
len=strlen(s);
sort(s,s+len,comp);
if(dfs())
{
for(int i=;i<;i++)
{
printf("%c",buf[i]+'A'-);
}
printf("%c\n",buf[]+'A'-);
}
else
{
printf("no solution\n");
}
}
return ;
}

HDOJ1015(简单深搜)的更多相关文章

  1. POJ 2386 Lake Counting (简单深搜)

    Description Due to recent rains, water has pooled in various places in Farmer John's field, which is ...

  2. 简单深搜:POJ1546——Sum it up

    结束了三分搜索的旅程 我开始迈入深搜的大坑.. 首先是一道比较基础的深搜题目(还是很难理解好么) POJ 1564 SUM IT UP 大体上的思路无非是通过深搜来进行穷举.匹配 为了能更好地理解深搜 ...

  3. poj 1562 简单深搜

    //搜八个方向即可 #include<stdio.h> #include<string.h> #define N 200 char ma[N][N]; int n,m,vis[ ...

  4. POJ-1321棋盘问题(简单深搜)

    简单搜索step1 POJ-1321 这是第一次博客,题目也很简单,主要是注意格式书写以及常见的快速输入输出和文件输入输出的格式. 递归的时候注意起始是从(-1,-1)开始,然后每次从下一行开始递归. ...

  5. NYoj The partial sum problem(简单深搜+优化)

    题目链接:http://acm.nyist.edu.cn/JudgeOnline/problem.php?pid=927 代码: #include <stdio.h> #include & ...

  6. nyoj587 hdu1045 简单深搜

    #include<iostream> #include<cstdio> #include<queue> #include<vector> #includ ...

  7. poj 2386:Lake Counting(简单DFS深搜)

    Lake Counting Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 18201   Accepted: 9192 De ...

  8. 【笔记】「pj复习」深搜——简单剪枝

    深搜--简单剪枝 说在最前面: 因为马上要 NOIP2020 了,所以菜鸡开始了复习qwq. pj 组 T1 ,T2 肯定要拿到满分的,然后 T3 , T4 拿部分分, T3 拿部分分最常见的做法就是 ...

  9. 【BZOJ】1016: [JSOI2008]最小生成树计数 深搜+并查集

    最小生成树计数 Description 现在给出了一个简单无向加权图.你不满足于求出这个图的最小生成树,而希望知道这个图中有多少个不同的最小生成树.(如果两颗最小生成树中至少有一条边不同,则这两个最小 ...

随机推荐

  1. Ubuntu中安装FTP 服务器自己踩得坑

    12点多了,擦!做个码农真不容易呀! 系统:Ubuntu16.04 安装:FTP 步骤: 1.不管有没有一上来我先卸载: sudo apt-get purge vsftpd 2.再安装:sudo ap ...

  2. 关于Ninja中上传下载文件

    上传得时候 根据类型来判断一下: 然后下载的时候需要:在url的最后一个/ 后面加上你要下载的类型如/download/app.apk; 下载成功就是app.apk了;

  3. springcloud zuul 使用zuulfilter 修改请求路径和响应头

    最近做项目有一个需求:一个网盘系统,文件存放在分布式文件系统中,之前的文件下载统一走的文件下载服务,现在需要在单文件下载的时候不需要走文件下载服务,而是直接访问文件系统上的路径,响应的时候修改响应头, ...

  4. READ_TEXT

    [转自http://lz357502668.blog.163.com/blog/static/1649674320109119101907/]这里,定义ITAB内表来存储长文本,并放到内表ITAB_E ...

  5. Python:笔记(1)——基础语法

    Python:笔记(1)——基础语法 我很抱歉有半年没有在博客园写过笔记了,客观因素有一些,但主观原因居多,再多的谴责和批判也都于事无补,我们能做的就是重振旗鼓,继续出发! ——写在Python之前 ...

  6. VHDL学习笔记——数字系统设计

    数字系统是指有若干数字电路和逻辑不见构成的能够处理或传输数字信息的设备.数字系统可分为三部分:输入输出接口.数据处理器和控制器. [传统的系统硬件设计]方法是(1)采用自底向上的设计方法(2)采用通用 ...

  7. 两个小例子彻底明白python decorator

    一:没有什么实际意思,就是单纯的理解decorator.使用装饰器完全可以阻止方法中的代码执行. class json_test(object): def __init__(self, *arg, * ...

  8. mini2440移植uboot 2014.04(四)

    我修改的代码已经上传到github上,地址:https://github.com/qiaoyuguo/u-boot-2014.04-mini2440.git 参考文章: <mini2440移植u ...

  9. Docker 数据管理-三种数据mount方式

    可以在Container可写层存储数据,但是有三个缺点: 当Container销毁时,数据不能持久保存. Container的可写层和Container所在的主机紧耦合,不容易将数据移动到其他地方. ...

  10. 利用Tkinter做的自动生成JSONSchema的小工具

    前面讲到可以使用JSONSchema做json数据校验, 但是每个接口数据都手动写jsonschema太痛苦了, 就写了个小脚本,可以直接复制接口文档的mock数据然后生成一个简单的jsonschem ...