P4379 [USACO18OPEN]Lemonade Line

最差情况就是耐性最好的牛都在前面.
照此贪心, 排个序以后看从哪个牛开始耐不住就OK
唯一要注意的就是所有牛的耐性都大于\(N\)的情况.

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int MAXN = 1e5 + 20;
inline int read()
{
    int x = 0; char ch = getchar();
    while(!isdigit(ch)) ch = getchar();
    while(isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
    return x;
}

int N;
int a[MAXN];

int main()
{
    cin>>N;
    for(int i = 1; i <= N; i++) a[i] = read();
    sort(a + 1, a + N + 1, greater<int>());
    int ans = N;
    for(int i = 1; i <= N; i++) if(a[i] < i - 1) {
        ans = i - 1; break;
    }
    cout<<ans<<endl;
    return 0;
}

P4379 [USACO18OPEN]Lemonade Line

原文：https://www.cnblogs.com/wsmrxc/p/9458704.html