1117. Eddington Number

1117. Eddington Number(25)

时间限制

250 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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:

106 7 6 9 3 10 8 2 7 8

Sample Output:

6

考试时候差点没看懂题，确实感觉描述的有点不清楚。E是说在N天中有E天骑行的路程超过E，然后求E的最大值，像是那种约束问题的说法。

排序一下，从E=N开始递减，每次对比数组中miles[N-E](E>=1)一个元素即可，E有可能为0。

#include<iostream>#include<algorithm>#include<vector>using namespace std;int main(){int n;cin>>n;vector<int>miles;for(int i=0;i<n;i++){int mile;cin>>mile;miles.push_back(mile);}sort(miles.begin(),miles.end());int Emax=n;while(Emax>=1&&miles[n-Emax]<=Emax)Emax--;cout<<Emax;return 0;}