POJ3617——Best Cow Line
来源:互联网 发布:苹果电脑网络不可用 编辑:程序博客网 时间:2024/06/05 11:10
Description
FJ is about to take his N (1 ≤ N ≤ 2,000) cows to the annual"Farmer of the Year" competition. In this contest every farmer arranges his cows in a line and herds them past the judges.
The contest organizers adopted a new registration scheme this year: simply register the initial letter of every cow in the order they will appear (i.e., If FJ takes Bessie, Sylvia, and Dora in that order he just registers BSD). After the registration phase ends, every group is judged in increasing lexicographic order according to the string of the initials of the cows' names.
FJ is very busy this year and has to hurry back to his farm, so he wants to be judged as early as possible. He decides to rearrange his cows, who have already lined up, before registering them.
FJ marks a location for a new line of the competing cows. He then proceeds to marshal the cows from the old line to the new one by repeatedly sending either the first or last cow in the (remainder of the) original line to the end of the new line. When he's finished, FJ takes his cows for registration in this new order.
Given the initial order of his cows, determine the least lexicographic string of initials he can make this way.
Input
* Line 1: A single integer: N
* Lines 2..N+1: Line i+1 contains a single initial ('A'..'Z') of the cow in the ith position in the original line
Output
The least lexicographic string he can make. Every line (except perhaps the last one) contains the initials of 80 cows ('A'..'Z') in the new line.
Sample Input
6ACDBCB
Sample Output
ABCBCD
Source
题意:
给定一个字符串,将其重新排列,组成一个字典序尽量小的字符串,重新排列的方式为:原字符串的两端字符比较,较小的一个放到新字符串的开头。
解:
贪心解决,两端开始扫描,找寻较小的字符,如果相同则比较下一个字母,当不同时输出较小一侧的字符。
#include <stdio.h>#include <string.h>#include <iostream>using namespace std;char str[2005];int main(){int n;while(~scanf("%d",&n)){int count=0;memset(str,0,sizeof(str));for(int i=0;i<n;i++){cin>>str[i];}int a,b;a=0;b=n-1;while(a<=b){int flag=0;for(int i=0;a+i<=b;i++){if(str[a+i]<str[b-i]){flag=1;break;}if(str[a+i]>str[b-i]){flag=0;break;}}if(flag)printf("%c",str[a++]);elseprintf("%c",str[b--]);count++;if(count%80==0)printf("\n");}}return 0;}
- POJ3617——Best Cow Line
- POJ3617 Best Cow Line
- POJ3617 Best Cow Line
- POJ3617 Best Cow Line
- best cow line(poj3617)
- POJ3617-Best Cow Line
- POJ3617 Best Cow Line
- POJ3617 Best Cow Line
- poj3617 Best Cow Line
- POJ3617 Best Cow Line
- POJ3617-Best Cow Line
- poj3617 Best Cow Line
- poj3617 Best Cow Line
- POJ3617 Best Cow Line
- POJ3617 Best Cow Line
- poj3617 Best Cow Line 贪心
- POJ3617 Best Cow Line 贪心
- poj3617 Best Cow Line 贪心
- 使用Vue框架实现NGA客户端
- 【JZOJ 3623】【SDOI2014】数表
- uva/6-12 572
- Linux下进程间通信方式——管道
- PAT-A1091
- POJ3617——Best Cow Line
- 要明确几个地方是结束不能占用的,
- 实例—ViewPager+RadioGroup实现底部导航栏和页面的滑动
- Codeforces Round #403 C. Andryusha and Colored Balloons
- 一起Talk Android吧(第十一回:Java中的继承)
- 520. Detect Capital
- POJ1001
- Vue上传图片预览以及删除的vue组件
- 有关计算机体系结构和人工智能的随想