1117 Eddington Number (25分)

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本文介绍了一种基于骑行距离计算Eddington数的方法，通过优化算法避免了常规方法的超时问题，使用数组记录每日骑行距离，实现了快速查找满足条件的最大整数E。

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:

题意：有N天的骑行距离，其中找个最大整数E（E<=N）满足有E天每天骑行距离均大于E。

思路：常规方法超时，可以用数组index来记录每个骑行距离有多少天，从1开始遍历，累加数组值，超过E距离的天数就等于【总天数 - 数组累加值（也即小于等于的天数）】若超过天数 <= E，保存E，若超过天数 > E，则结束循环，输出保存的最后一个E即可。（逻辑有点儿绕，静下心来好好想想～）

#include<iostream>
#include<vector>


using namespace std;


int main(){

    int n;
    cin >> n;

    int a[100010] = {0};

    for(int i = 0; i < n; i++){
        int temp;
        scanf("%d", &temp);
        if(temp <= 100000 )
            a[temp]++;

    }
    int flag = 0;
    if(n == 0){
        printf("0\n");
        return 0;

    }else
    {
        int cnt = 0;
        
        for(int i = 1; i <=n; i++){
            cnt += a[i];
            if(i <= n - cnt){
                flag = i;
            }else
            {
                printf("%d\n", flag);
                return 0;
            }
            

        }
    }
    printf("%d\n", flag);
    
    
    return 0;
}