POJ3617

Best Cow Line

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：给定一个字符串，重排，每次从头或尾取较小值放到新序列的末尾。
需要注意的地方：不能简单的每次首尾比较取最小值，因为如果相等还需要继续看下一个要比较的对儿那个更小。
题目要求是每80个为一行输出。

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
char a[2001];
int n;
int main()
{
    int i, f, l, cnt;
    int left;
    while (scanf("%d",&n)!=EOF){
        //getchar();
        for (i = 0; i < n; i++){
            char c;    //起缓冲作用，好神奇的东东（再也不用担心读入空格和换行了。。。）
            cin >> c;
            a[i] = c;
        }
        cnt = 0;
        f = 0; l = n - 1;
        while (f <= l){
            left = 0;
            for (i = 0; f + i <= l; i++){　　//只要你看懂了这个，这道题的精髓你以已经掌握了
                if (a[f + i] < a[l - i]){
                    left = 1;
                    break;
                }
                else if (a[f + i]>a[l - i]){
                    left = 0;
                    break;
                }
            }
            if (left){
                cout<<a[f];
                f += 1;
            }
            else{
                cout<<a[l];
                l -= 1;
            }
            cnt += 1;
            if (cnt % 80 == 0){ cout<<"
"; }
        }
    }
    
    //system("pause");
    return 0;
}

 重新刷题，感觉曾经的自己是那么的稚嫩。

#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#define MAX 2000
using namespace std;

char cow[MAX];

void solve(){
    int n, l, r, line = 0;
    bool left;
    cin >> n;
    for (int i = 0; i < n; i++){
        cin >> cow[i];
    }
    l = 0; r = n - 1;
    while (l <= r){
        left = true;
        for (int i = 0; l + i <= r; i++){
            if (cow[l + i] < cow[r - i])break;
            else if (cow[l + i]>cow[r - i]){ left = false; break; }
        }
        if (left)putchar(cow[l++]);
        else putchar(cow[r--]);
        line++;
        if (line == 80){ line = 0; putchar('
'); }
    }
    putchar('
');
}

int main()
{
    solve();
    //system("pause");
    return 0;
}