PAT (Advanced Level) Practise 1117 Eddington Number(25)

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

题意：在N天中有E天骑行的路程超过E，然后求E的最大值

解题思路：对n个数据从大到小排序，i表示了骑车的天数，那么满足a[i] > i的最大值即为所求

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <functional>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;

int main()
{
int n,a[100005];
while(~scanf("%d",&n))
{
for(int i=1;i<=n;i++) scanf("%d",&a[i]);
sort(a+1,a+1+n,greater<int>());
int k=1;
while(a[k]>k) k++;
printf("%d\n",k-1);
}
return 0;
}