1117. Eddington Number(25)
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
找出一个E,有E天,行程超过E miles,先从小到大排序,从后往前看,如果当前看了s个数,而最后一个数又比s大,那么s满足条件,找出最大的s。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int s[];
int n,m;
int main()
{
cin>>n;
for(int i = ;i < n;i ++)
{
cin>>s[i];
}
sort(s,s + n);
for(int i = n - ;i >= ;i --)
{
if(s[i] > n - i && m < n - i)m = n - i;
}
cout<<m;
}
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