PAT_A1117#Eddington Number

Source:

PAT A1117 Eddington Number (25 分)

Description:

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

Keys:

• 模拟题

Code:

 1 /*
 2 Data: 2019-07-19 18:27:10
 3 Problem: PAT_A1117#Eddington Number
 4 AC: 13:54
 5 
 6 题目大意：
 7 E数定义：E天之中，骑行米数超过E的最大天数
 8 
 9 基本思路：
10 从大到小排序，找到最大的i，使A[i]>i
11 */
12 
13 #include<cstdio>
14 #include<functional>
15 #include<algorithm>
16 using namespace std;
17 const int M=1e5+10;
18 int e[M];
19 
20 int main()
21 {
22 #ifdef    ONLINE_JUDGE
23 #else
24     freopen("Test.txt", "r", stdin);
25 #endif
26 
27     int n,pos;
28     scanf("%d", &n);
29     for(int i=1; i<=n; i++)
30         scanf("%d", &e[i]);
31     sort(e+1,e+n+1,greater<int>());
32     for(pos=1; pos<=n; pos++)
33         if(e[pos] <= pos)
34             break;
35     if(pos==n && e[n]>n)
36         printf("%d", pos);
37     else
38         printf("%d", pos-1);
39 
40     return 0;
41 }

PAT_A1117#Eddington Number

原文：https://www.cnblogs.com/blue-lin/p/11215169.html