Safecracker

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13237    Accepted Submission(s): 6897

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<iostream>
#include<cstdio>
#include<cmath>
#include<map>
#include<cstdlib>
#include<vector>
#include<set>
#include<queue>
#include<cstring>
#include<string.h>
#include<algorithm>
#define INF 0x3f3f3f3f
typedef long long ll;
typedef unsigned long long LL;
using namespace std;
const int N = 1e6+;
const ll mod = 1e9+;
char s[];
int a[];
int b[];
int visited[];
int n;
int flag=;
void DFS(int t){
if(flag==)return;
if(t==){
int sum1=b[]+b[]*b[]*b[]+b[]*b[]*b[]*b[]*b[];
int sum2=b[]*b[]+b[]*b[]*b[]*b[];
if(sum1-sum2==n){
flag=;
printf("%c%c%c%c%c\n",b[]-+'A',b[]-+'A',b[]-+'A',b[]-+'A',b[]-+'A');
return ;
}
return ;
}
for(int i=;i>=;i--){
if(visited[i]==)continue;
else if(visited[i]==&&a[i]==){
visited[i]=;
b[t]=i;
DFS(t+);
visited[i]=;
}
}
}
int main(){
while(scanf("%d %s",&n,s)!=EOF){
if(n==&&strcmp(s,"END")==)break;
int len=strlen(s);
flag=;
memset(a,,sizeof(a));
for(int i=;i<len;i++){
int tt=s[i]-'A'+;
a[tt]=;
}
memset(visited,,sizeof(visited));
DFS();
if(flag==)cout<<"no solution"<<endl; } }

hdu 1015(DFS)的更多相关文章

  1. HDOJ(HDU).1015 Safecracker (DFS)

    HDOJ(HDU).1015 Safecracker [从零开始DFS(2)] 从零开始DFS HDOJ.1342 Lotto [从零开始DFS(0)] - DFS思想与框架/双重DFS HDOJ.1 ...

  2. hdu 1342(DFS)

    Lotto Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submi ...

  3. F - JDG HDU - 2112 (最短路)&& E - IGNB HDU - 1242 (dfs)

    经过锦囊相助,海东集团终于度过了危机,从此,HDU的发展就一直顺风顺水,到了2050年,集团已经相当规模了,据说进入了钱江肉丝经济开发区500强.这时候,XHD夫妇也退居了二线,并在风景秀美的诸暨市浬 ...

  4. F - Auxiliary Set HDU - 5927 (dfs判断lca)

    题目链接: F - Auxiliary Set HDU - 5927 学习网址:https://blog.csdn.net/yiqzq/article/details/81952369题目大意一棵节点 ...

  5. hdu - 1072(dfs剪枝或bfs)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1072 思路:深搜每一个节点,并且进行剪枝,记录每一步上一次的s1,s2:如果之前走过的时间小于这一次, ...

  6. HDU——2647Reward(DFS或差分约束)

    Reward Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Sub ...

  7. P - Sudoku Killer HDU - 1426(dfs + map统计数据)

    P - Sudoku Killer HDU - 1426 自从2006年3月10日至11日的首届数独世界锦标赛以后,数独这项游戏越来越受到人们的喜爱和重视. 据说,在2008北京奥运会上,会将数独列为 ...

  8. hdu 1142(DFS+dijkstra)

    #include<iostream> #include<cstdio> #include<cmath> #include<map> #include&l ...

  9. hdu 1181(DFS)变 形 课

    变形课 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)Total Submis ...

随机推荐

  1. jquery 获取form下的所有元素

    <!DOCTYPE html> <html> <head> <meta charset="gb2312"> <title> ...

  2. add number

    // io.cpp #include <iostream> int readNumber() { std::cout << "Enter a number: &quo ...

  3. Rust的几个预测

    写程序多年,语言也用过不下十种,对于Rust有种亲人的感觉,就像在梦中见到过似的.现在对于Rust特做出以下一些预测,希望Rust会有更大的影响力. 1. 当前的Rust的核心功能现以比较稳定,可以用 ...

  4. java 访问mysql 实例

    前提条件: 1.安装eclipse,mysql.java jdk 2.安装mysql connect J  (我安装的版本是mysql connect J 5.1.39) 3.配置java环境变量 4 ...

  5. Js文件中文乱码

    aspx页面引用的js文件中如果包括中文,中文显示乱码或者引起脚本错误.提示是'未结束的字符串' 原因:aspx页面的默认编码是utf-8,而js文件的默认编码是gb2312,两者之间不一致引起了中文 ...

  6. tcp转发

    Proxy.java package com.dc.tcp.proxy; import java.io.IOException; import java.net.ServerSocket; impor ...

  7. Observer(观察者)-对象行为型模式

    1.意图 定义对象间的一种一对多的依赖关系,当一个对象的状态发生改变时,所有依赖于它的对象都得到通知并被自动更新. 2.别名 依赖(Depenents),发布-订阅(Publish-subscribe ...

  8. python第十九天-----Django进阶

    1.机智的小django为我你们提供了快捷的表单验证! from django.shortcuts import render, HttpResponse,redirect from django i ...

  9. 转:union和union all的区别

    Union因为要进行重复值扫描,所以效率低.如果合并没有刻意要删除重复行,那么就使用Union All  两个要联合的SQL语句 字段个数必须一样,而且字段类型要“相容”(一致): 如果我们需要将两个 ...

  10. C# 通过代理获取url数据

    public static string doPost(string Url, byte[] postData, SinaCookie bCookie, String encodingFormat, ...