A1117 Eddington Number (25 point(s))

简单推理

1. 原文

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 Emiles. 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

2. 解析

从大到小排序，记录个数，到值<个数时，找到该值

3. AC代码

#include<cstdio>
#include<algorithm>
using namespace std;
int num[100010]={};
bool cmp(int x,int y){
	return x>y;
}
int main()
{
	int n;
	scanf("%d",&n);
	for (int i = 0; i < n; ++i)
	{
		scanf("%d",&num[i]);
	}
	sort(num,num+n,cmp);
	int ans=1;
	for (int i = 0; i < n; ++i)
	{
		if (num[i]>ans)
		{
			ans++;
		}
	}
	printf("%d\n",ans-1);
	return 0;
}