PAT A1117 Eddington Number （25 分）——数学题

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 (≤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

 

#include <stdio.h>
#include <algorithm>
#include <iostream>
#include <map>
#include <vector>
#include <queue>
#include <set>
using namespace std;

const int maxn=100010;
int dis[maxn];

int main(){
    int n;
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        int id;
        scanf("%d",&id);
        if(id>100000)id=100001;
        dis[id]++;
    }
    int cnt=0;
    int i;
    for(i=100009;i>=0;i--){
        if(cnt>=i)break;
        cnt+=dis[i];
    }
    printf("%d",i);
}

注意点：其实就是找最大的满足条件的数，条件就是大于指定数的个数是否大于这个数。但是注意从后往前遍历增加时，不能输出大于它的个数，要输出指定数，还有超过100000的数要特殊处理