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
- 分析
待续 - code
#include <iostream>
#include <algorithm>
#include <deque>
using namespace std;
int main()
{deque<int> E;int N;cin>>N;int tmp;for(int i=0;i<N;i++){cin>>tmp;E.push_back(tmp);}sort(E.begin(),E.begin()+N);int E1=0;for(int i=0;i<N;i++){tmp=E.at(0);E.pop_front();for(int j=E1;j<tmp;j++){if(j==N-i)E1=j;else if(j>N-i){break;}}}cout<<E1<<endl;return 0;
}