Best Cow Line POJ - 3617 (贪心)

https://vjudge.net/problem/POJ-3617

很简单的贪心水题,

(2) 如果相等, 再向后/前, 判断(1)

(1)每次都取两边最小的

注意仔细读取, 题中还有一个要求每输够80次多输出一个换行, 所以说英文真的很重要

#include <iostream>
#include <cstdio>
#include <string>
#include <algorithm>
using namespace std;

const int maxn = 2010;
char cows[maxn];
int N;
void solve()
{
    int a = 0, b = N-1, ans = 0;
    while(a <= b){
        bool left = false; //左边是否小于右边
        for(int i = 0; a+i <= b; i++){ //如果左右相等, 一直找直到找到一个能比较大小的后/前字母
            if (cows[a+i] < cows[b-i]){
                left = 1;
                break;
            }
            else if (cows[a+i] > cows[b-i]){
                left = 0;
                break;
            }
        }
        if (left) printf("%c",cows[a++]);
        else printf("%c",cows[b--]);
        ans++;
        if (ans == 80) printf("\n"), ans = 0;
    }
    printf("\n");
}
int main() {
    cin >> N;
    for(int i = 0; i < N; i++){
        scanf(" %c",&cows[i]);
    }
    solve();
    return 0;
}