Eddington, a British astronomer, likes to ride bicycles. It is said that in order to show off his cycling skills, he also defined an "Edington number" E, that is, the largest integer E that satisfies E days of cycling for more than E miles. Eddington's own E is said to be equal to 87.
Now, given someone's cycling distance in N days, please calculate the corresponding Eddington number E (<=N).
Input format:
输入第一行给出一个正整数N（<=105），即连续骑车的天数；第二行给出N个非负整数，代表每天的骑车距离。
Output format:
在一行中给出N天的爱丁顿数。
Input sample:
10
6 7 6 9 3 10 8 2 7 8
Sample output:
6
Problem- solving idea: first sort the array from large to small, and regard someone's cycling distance as decreasing every day. a[i] is the riding distance on the i-th day. If a[i]>i, it means that there are i days where the riding distance is greater than a[i] until a[i]
#include <algorithm>
using namespace std;
int a[1000000];
bool cmp(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, cmp);
int ans = 0, p = 1;
while(ans <= n && a[p] > p) {
ans++;
p++;
}
printf("%d", ans);
return 0;
}