PAT A1117 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:

 #include <stdio.h>

 #include <algorithm>

 #include <iostream>

 #include <map>

 #include <vector>

 #include <queue>

 #include <set>

 using namespace std;

 const int maxn=;

 int dis[maxn];

 int main(){

     int n;

     scanf("%d",&n);

     for(int i=;i<=n;i++){

         int id;

         scanf("%d",&id);

         if(id>)id=;

         dis[id]++;

     }

     int cnt=;

     int i;

     for(i=;i>=;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

巴特西

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

Input Specification:

Output Specification:

Sample Input:

Sample Output:

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