PAT甲级1117

1117. Eddington Number(25)

时间限制

250 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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(<=10⁵), 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

#include <cstdio>
#include <algorithm>
using namespace std;
int a[1000000];
bool cmp1(int a, int b) {
  return a > b;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  sort(a + 1, a + n + 1, cmp1);//从大到小排序
  int ans = 0;
  int p = 1;
  while (ans <= n && a[p] > p) {//若满足，那么1~p-1对应的数肯定是大于p的，然后如果a[p]>p那么刚好就满足那个爱丁顿数
    ans++;
    p++;
  }
  printf("%d", ans);
  return 0;
}