1117 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 (≤).

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (≤), 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

别的题目我都很少写blog，大部分总结都是写在notebook上，但我觉得这个题目代码思路我写的很好，复杂度比系统自带的sort函数应该要低，能达到o(n),可以和大家分享一下。

#include <iostream>
using namespace std;
const int maxsize=100001;
int a[maxsize],n;
bool is_ok(int s){//遍历数组判断是否满足大于s的天数大于等于s
    int counts=0;
    for(int i=0;i<n;i++){
        if(a[i]>s)counts++;
    }
    if(counts>=s)return true;
    else return false;
}
void find_ee(int low,int high,int& en){
    if(low>high)return;
    int mid=(low+high)/2;
    if(is_ok(mid)){        
        en=mid;//放在前面递归结束时不会修改en        
        find_ee(mid+1,high,en);
    }else{
        find_ee(low,mid-1,en);
    }
}
int main(){
    cin>>n;
    int en=0;
    for(int i=0;i<n;i++)cin>>a[i];
    find_ee(1,n,en);
    cout<<en;
    return 0;
}