1117. Eddington Number(25)

最新推荐文章于 2023-01-13 09:03:36 发布

原创最新推荐文章于 2023-01-13 09:03:36 发布 · 223 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#PAT #PAT甲

PAT 甲级专栏收录该内容

97 篇文章

订阅专栏

本文介绍了一个关于自行车骑行里程的算法问题——EddingtonNumber。该问题要求找到连续骑行天数中，超过某特定距离的最多天数。文章通过一个示例详细解释了解决方案，并提供了AC代码。

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:

10
6 7 6 9 3 10 8 2 7 8

Sample Output:

----------------------------------------------------------

“E -- that is, the maximum integer E such that it is for E days that one rides more than E miles.”

个人感觉读懂这个长难句比做题目本身还难.....

其实题目很简单，那例子来说，输出6的理由是：这N个数里有6个数比6大，所以E=6..（且E要尽可能地大）

AC代码：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <vector>
#include <queue>
#include <stack>

using namespace std;


int main()
{
    freopen("in.txt","r",stdin);

    int N,N1;
    cin>>N;
    N1=N;
    vector<int> a;

    while(N1--)
    {
        int t;
        cin>>t;
        a.push_back(t);
    }
    sort(a.begin(),a.end());
    int i;
    for(i=N;i>=1;i--)
    {
        if(i<a[N-i])
            break;
    }
    printf("%d",i);

    return 0;
}