PAT - 甲级 - 1117. Eddington Number(25) (题意理解)

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

题目大意：给出n天的骑车距离，求出爱丁顿数，其实就是求出E，这个E表示E天的骑行距离超过E英里。

这里需要理解的超过是大于还是大于等于，题目中说you are supposed to find the corresponding E (<=N).  说明超过的意思是大于。

思路：对给出的骑行距离排序后，从大到小扫描一遍就是答案。

证明：骑行距离从大到小排列后，假如第三天的距离小于3，再次进行扫描的时候，第四天的距离比3还小，肯定不会大于4，第五天的数据更小，同理，之后的越来越小，因此2就是答案，这个2不会更大。

#include<cstdio>
#include<algorithm>
#define N 100001
using namespace std;
int j,n,ans,a
;
int main(){
scanf("%d",&n);
//数据输入
for(int i=0 ;i<n ;i++) scanf("%d",&a[i]);

sort(a,a+n);
//数据处理
for(j=n-1,ans=1 ;j>=0 ;j--,ans++){
if(a[j]<=ans)break;
}
//因为当判断到<=才退出循环，因此答案是前一天
printf("%d\n",ans-1);

return 0;
}