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

题意：三个E  。最大的E满足有E天大于E公里数这个条件的。 先将数组按小到大排序   从大到小更方便，

ac代码：#include <iostream>
#include <map>
#include <cstdio>
#include <vector>
#include <string>
#include <algorithm>
using namespace std;

int num[100005];
int main(){

//freopen("E:\input.txt","r",stdin);
int n;
cin>>n;
for(int i=0;i<n;i++) scanf("%d",&num[i]);

sort(num,num+n);
//for(int i=0;i<n;i++) printf("%d ",num[i]);
//cout<<endl;
int ans=num[0]-1;

for(int i=0;i<n;){
//if(num[i+1]==num[i]) i++;
//else{
if(n-i>=num[i]-1) {//printf("i:%d ans:%d cnt:%d\n",i,num[i]-1,n-i);
ans=num[i]-1; if(i<n-1) while(num[i+1]==num[i]) {i++; if(i>=n-1) break;}
i++;}
else break;
//}
}

cout<<ans;//<<endl;
freopen("CON","r",stdin);
system("pause");
}