PAT甲级 1117 Eddington Number （25 分）逻辑题

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本文介绍了一种基于骑行距离的挑战——EddingtonNumber，通过算法找出连续骑行天数中，超过特定距离的天数最大值。输入连续N天的骑行距离，输出满足条件的最大整数E。

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1117 Eddington Number （25 分）

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:

题目大意：找出一个数e，满足该数字串中大于e的数大于等于e个。

解题思想：从大到小排序后，从最大的数开始遍历，e初始为0，判断e还能继续增加，只要判断此时v[i]是否大于e+1；这样做就满足了题意。如果换成暴力求解会有一个点超时。

#include<iostream>
#include<algorithm> 
#include<cstdio>
#include<vector>
using namespace std;

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

int main(){
	int n;
	cin>>n;
	vector<int> vec(n);
	for(int i=0;i<n;i++)
		scanf("%d",&vec[i]);
	sort(vec.begin(),vec.end(),cmp);
	int e = 0;
	for(int i=0;i<n;i++){
		if(vec[i] > e+1)
			e++;
		else
			break;
	}
	cout<<e;
	
	return 0;
}