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:
6
1 #include <stdio.h> 2 #include <algorithm> 3 #include <iostream> 4 #include <map> 5 #include <vector> 6 #include <queue> 7 #include <set> 8 using namespace std; 9 10 const int maxn=100010; 11 int dis[maxn]; 12 13 int main(){ 14 int n; 15 scanf("%d",&n); 16 for(int i=1;i<=n;i++){ 17 int id; 18 scanf("%d",&id); 19 if(id>100000)id=100001; 20 dis[id]++; 21 } 22 int cnt=0; 23 int i; 24 for(i=100009;i>=0;i--){ 25 if(cnt>=i)break; 26 cnt+=dis[i]; 27 } 28 printf("%d",i); 29 }
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