【笨方法学PAT】1117 Eddington Number （25 分）_pat 1117 eddington number(25 分)-优快云博客

本文链接：https://blog.youkuaiyun.com/linghugoolge/article/details/84664621

本文介绍了一种基于数组和排序算法的解决方案，用于计算Eddington数，即找到连续N天骑行距离中，超过E天的E英里最大整数值E。通过输入连续骑行天数和每天的骑行距离，程序能够输出相应的Eddington数。

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一、题目

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:

二、题目大意

给一串数字，求大于E的E个数的最大E值。

三、考点

数组

四、注意

1、理解题意。

五、代码

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
bool cmp(int a, int b) {
	return a > b;
}
int main() {
	//read
	int n;
	cin >> n;
	vector<int> v(n);
	for (int i = 0; i < n; ++i) {
		cin >> v[i];
	}

	//sort
	sort(v.begin(), v.end(), cmp);

	int i;
	for (i = 0; i < n; ++i) {
		if (v[i] <= i + 1)
			break;
	}

	cout << i << endl;

	system("pause");
	return 0;
}