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

題目大意：給定一個(gè)字符串，重排，每次從頭或尾取較小值放到新序列的末尾。
需要注意的地方：不能簡(jiǎn)單的每次首尾比較取最小值，因?yàn)槿绻嗟冗€需要繼續(xù)看下一個(gè)要比較的對(duì)兒那個(gè)更小。
題目要求是每80個(gè)為一行輸出。 #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; //起緩沖作用，好神奇的東東（再也不用擔(dān)心讀入空格和換行了。。。）cin >> c;a[i] = c;}cnt = 0;f = 0; l = n - 1;while (f <= l){left = 0;for (i = 0; f + i <= l; i++){　　//只要你看懂了這個(gè)，這道題的精髓你以已經(jīng)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<<"\n"; }}}//system("pause");return 0; }

?重新刷題，感覺(jué)曾經(jīng)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('\n'); }}putchar('\n'); }int main() {solve();//system("pause");return 0; }

?

轉(zhuǎn)載于:https://www.cnblogs.com/littlehoom/p/3550281.html

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