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
题目大意是给定一个长度为N的字符串S，要求构造一个字符串T。T的默认状态为空。然后随机将S的头部（或尾部）删除一个字符，加到T的尾部。
要求构造一个字典序尽可能小的字符串T。
思路：
- 按照字典序比较S和反转后的字符串S’
- 如果S较小，就从S开头取出字符，放到T的末尾
- 如果S’较小，就从S末位取出字符，放到T的末尾
- 如果相同，则比较(S+1)和(S’-1)两字符的大小；相同则继续。
字典序+贪心问题。
出现PE问题：每行最多80个字符
#include
#include
#include
int N,i;
const int MAX_N = 2000;
char S[MAX_N + 1];
int main()
{
while (~scanf("%d", &N))
{
for (i = 0; i < N; i++)
{
getchar();
scanf("%c", &S[i]);
}
int a = 0, b = N - 1, cnt;
cnt = 0;
while (a <= b)
{
bool left = false;//比较左侧和右侧的字符串
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)putchar(S[a++]);
else putchar(S[b--]);
cnt++;
if (cnt % 80 == 0)
putchar('n');
}
}
return 0;
}
