PAT 甲级 1117 Eddington Number

https://pintia.cn/problem-sets/994805342720868352/problems/994805354762715136

British astronomer Eddington liked to ride a bike. It is said that in order to show off his skill, he has even defined an "Eddington number", E -- that is, the maximum integer E such that it is for E days that one rides more than E miles. Eddington's own E was 87.

Now given everyday's distances that one rides for N days, you are supposed to find the corresponding E (≤N).

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (≤10​5​​), the days of continuous riding. Then N non-negative integers are given in the next line, being the riding distances of everyday.

Output Specification:

For each case, print in a line the Eddington number for these N days.

Sample Input:

10
6 7 6 9 3 10 8 2 7 8

Sample Output:

6

 

代码：

#include <bits/stdc++.h>
using namespace std;
 
const int maxn = 1e5 + 10;
long long a[maxn];
 
int main() {
    int n;
    scanf("%d", &n);
    for(int i = 1; i <= n; i ++)
        scanf("%lld", &a[i]);
    sort(a + 1, a + 1 + n);
 
 
        for(int i = n; i >= 1; i --) {
            int L = 1, R = n, pos = -1;
            while(L <= R) {
                int mid = (L + R) / 2;
                if(a[mid] > 1LL * i) R = mid - 1, pos = mid;
                else L = mid + 1;
            }
 
            if(pos == -1) continue;
            if(n - pos + 1 >= i) {
                printf("%d
", i);
                return 0;
            }
        }
        printf("0
");
 
    return 0;
}

　　午睡结束 最近也太爱睡觉了吧