UVA 10132 - File Fragmentation
来源:互联网 发布:ssd固态硬盘优化软件 编辑:程序博客网 时间:2024/06/05 17:08
File Fragmentation
The Problem
Your friend, a biochemistry major, tripped while carrying a tray of computer files through the lab. All of the files fell to the ground and broke. Your friend picked up all the file fragments and called you to ask for help putting them back together again.
Fortunately, all of the files on the tray were identical, all of them broke into exactly two fragments, and all of the file fragments were found. Unfortunately, the files didn't all break in the same place, and the fragments were completely mixed up by their fall to the floor.
You've translated the original binary fragments into strings of ASCII 1's and 0's, and you're planning to write a program to determine the bit pattern the files contained.
Input
The input begins with a single positive integer on a line by itself indicating the number of the cases following, each of them as described below. This line is followed by a blank line, and there is also a blank line between two consecutive inputs.
Input will consist of a sequence of ``file fragments'', one per line, terminated by the end-of-file marker. Each fragment consists of a string of ASCII 1's and 0's.
Output
For each test case, the output must follow the description below. The outputs of two consecutive cases will be separated by a blank line.Output is a single line of ASCII 1's and 0's giving the bit pattern of the original files. If there are 2N fragments in the input, it should be possible to concatenate these fragments together in pairs to make N copies of the output string. If there is no unique solution, any of the possible solutions may be output.
Your friend is certain that there were no more than 144 files on the tray, and that the files were all less than 256 bytes in size.
Sample Input
1011011101110111011110111
Sample Output
01110111
================================
每次输入完毕后都会再输入一个空行……每组输出之间也要有空行……
先排序,总长度就是a[0]和a[n-1]的长度和。
长度相等的最多有两个(不会有两个文件在同一个地方断掉),所以只要出现有长度相等的ab,和与它们对应的长度相等的两个cd一定可以组成两份相同的原文件。只有ac ad bc bd四种方式,枚举即可。
如果都只出现一次,那么a[0]和a[n-1]、a[1]和a[n-2]一定可以组成两份相同的原文件,同样枚举即可。
#include <iostream>#include <string>#include <cstring>#include <cstdio>#include <algorithm>using namespace std;string str[333],s1,s2;char s[333];int cmp(string a,string b){ if(a.length()==b.length()) return a<b; else return a.length()<b.length();}int main(){ int t; scanf("%d",&t); getchar(); getchar(); while(t--) { int n=0,p=-1; while(gets(s)) { if(s[0]!='\0') str[n++]=s; else break; } sort(str,str+n,cmp); //for(int i=0;i<n;i++) cout<<"---"<<str[i]<<endl; int sumlen=str[0].length()+str[n-1].length(); for(int i=0;i<n-1;i++) { if(str[i].length()==str[i+1].length()) p=i; break; } if(p!=-1) { if(str[p]+str[n-1-p]==str[p+1]+str[n-2-p]) cout<<str[p]+str[n-1-p]<<endl; else if(str[p]+str[n-1-p]==str[n-2-p]+str[p+1]) cout<<str[p]+str[n-1-p]<<endl; else if(str[n-1-p]+str[p]==str[p+1]+str[n-2-p]) cout<<str[n-1-p]+str[p]<<endl; else if(str[n-1-p]+str[p]==str[n-2-p]+str[p+1]) cout<<str[n-1-p]+str[p]<<endl; } else { if(str[0]+str[n-1]==str[1]+str[n-2]) cout<<str[0]+str[n-1]<<endl; else if(str[0]+str[n-1]==str[n-2]+str[1]) cout<<str[0]+str[n-1]<<endl; else if(str[n-1]+str[0]==str[1]+str[n-2]) cout<<str[n-1]+str[0]<<endl; else if(str[n-1]+str[0]==str[n-2]+str[1]) cout<<str[n-1]+str[0]<<endl; } if(t) cout<<endl; } return 0;}
- UVa 10132 - File Fragmentation
- UVa 10132 - File Fragmentation
- UVa 10132 - File Fragmentation
- uva 10132 - File Fragmentation
- uva 10132 - File Fragmentation
- UVa 10132 - File Fragmentation
- UVa 10132 - File Fragmentation
- UVA 10132 File Fragmentation
- UVa:10132 File Fragmentation
- UVA 10132 - File Fragmentation
- uva 10132 File Fragmentation
- uva 10132 File Fragmentation
- uva 10132 File Fragmentation
- uva 10132 - File Fragmentation
- uva 10132 - File Fragmentation
- UVA - 10132 File Fragmentation
- uva 10132 File Fragmentation
- UVA 10132 File Fragmentation
- Android 权限大全
- JNA java调用c/c++代码
- MFC状态栏的编程
- 柠檬水的10个好处
- hdu 1032
- UVA 10132 - File Fragmentation
- jni入门之javah自动生成jni的c文件所需头文件
- Boost智能指针——weak_ptr
- Andriod UI设计之度量单位说明(DIP,DP,PX,SP)
- uva 11404 Palindromic Subsequence(LCS回文串,最小字典序)
- 关注分离
- 用51单片机设计一款记忆力测试小游戏
- cocosbuilder Version 3 alpha5的使用
- C#中char[]与string之间的转换