1117 Eddington Number (25 point(s))

1117 Eddington Number (25 point(s))

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 (≤10​5​​), 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

H-Index算法。

最开始的思路是哈希思想，但是如果有数字大于1e5就不可以了。

仔细一看，这就是H-Index的求法，即H篇被引量超过N的论文。

首先降序排序，以上面的例子来看：10 9 8 8 7 7 6 6 3 2。以下标为1位置为例，表示有1篇被引超过9的论文；以下标4为例，表示有2篇被引超过7的论文，但也可以表示成有3篇被引大于6（7-1）的论文，这样到这个下标位置H至少是3了。只要当前的篇数不大于当前最低的下界（不含），就可以循环继续。注意在下标i的位置（0<=i<N）是对前i个元素进行的判断，因此最后还需要判断H是否可以是N。

相关链接：https://blog.youkuaiyun.com/richenyunqi/article/details/79764568

#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
const int MAX = 1e5+7; 
int a[MAX]={0};
bool comp(const int &a,const int &b)
{
    return a>b;
}
int main(void){
	int N;cin>>N;
	for(int i=0;i<N;i++) cin>>a[i];
	sort(a,a+N,comp);
	int cnt;
	int ans =0; int max;
	for(int i=0;i<N;i++){
		if(i==0||a[i]!=a[i-1]){
			cnt=i;
			max = a[i]; 
		}//前面有cnt个元素大于a[i]的元素 
		else if(a[i]==a[i-1]){
			cnt = i;
			max = a[i]-1;
		}//前面有cnt个大于a[i-1]的元素 
		if(cnt<=max){//有cnt个大于a[i]的元素 
			ans=cnt;
		}
		else break;
	}
	if(a[N-1]>N) ans=N; 
	cout<<ans<<endl;
	return 0;
}