| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 8579 | Accepted: 2629 |
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
6 A C D B C B
Sample Output
ABCBCD
Source
#include <stdio.h>
#include <string.h>
char s[2002];
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
for(int i=0;i<n;i++)
scanf("\n%c",&s[i]);//这个地方参考了别人的，感觉好巧妙的，学习了，避免了输入的时候把回车保存的情况。（开始就是这个地方一直 // wa,不能一行输入，还是要看清题意。
int a=0,b=n-1;
int ans=0; //先前没注意，把这个语句定义到下面的while循环里面了，导致一直pe。。
while(a<=b) //首位字符进行比较
{
bool left=false;
for(int i=0;a+i<=b;i++)
{
if(s[a+i]<s[b-i])
{
left=true;
break;
}
else if(s[a+i]>s[b-i])
{
left=false;
break;
}
}
if(left) printf("%c",s[a++]);
else printf("%c",s[b--]);
ans++;
if(ans==80){ //这个地方我测试了一下 如果写成下面的样子的话，时间就是16ms
printf("\n");
ans=0;
}
/*if(ans==80)//这样写就是16ms，上面就是0ms，不知道是判题的问题，还是本身就节约时间，不解
{
printf("\n");
ans=0;
}*/
}
printf("\n");
}
return 0;
}一个这样的题目都还ac我好久啊，基础还是不扎实，自己还是要加强训练！！！加油！！
原文地址：http://blog.csdn.net/whjkm/article/details/24635219
