PAT A1117. Eddington Number(25)

1117. Eddington Number(25)

时间限制

250 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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:

106 7 6 9 3 10 8 2 7 8

Sample Output:

6

思路分析：

讲每日的距离存在数组ride中按从大到小排序，然后从i=1开始遍历，同时用k记录遍历过的个数，若遇到i >= ride[i]，停止，k = i-1。k即为结果

题解：

#include <cstdio>#include <algorithm>using namespace std;const int MAX = 100010;int ride[MAX] = {0};int n;bool cmp(int a, int b){return a > b;}int main(){scanf("%d", &n);for(int i = 1; i <= n; i++){scanf("%d", &ride[i]);}sort(ride+1, ride+n+1, cmp);int k = 0;for(int i = 1; i <= n; i++){if(i >= ride[i]){k = i-1;break;}k++;}printf("%d", k);return 0;}