1117. Eddington Number(25)

最新推荐文章于 2020-07-18 18:16:47 发布

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最新推荐文章于 2020-07-18 18:16:47 发布

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分类专栏： PAT 文章标签： PAT 1117

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这是一篇关于Eddington数的文章，介绍了英国天文学家Eddington定义的Eddington数，即一个人骑行天数大于骑行英里数的最大整数E。题目提供了一组骑行天数和每天的距离，要求找出对应的Eddington数。

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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:

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <iostream>

using namespace std;

typedef long long int ll;

bool cmp(int a,int b){
	return a>b;
}

int main(){
	
	int a[100005];
	int N;
	scanf("%d",&N);
	for(int i=1; i<=N; i++)
		scanf("%d",&a[i]);
	sort(a+1,a+1+N,cmp);
	int index;
	for(index = 1; a[index]>index; index++);
	cout << index-1 << endl;
	
	return 0;
}