Hdu6066 RXD's date(2017多校第3场)

最新推荐文章于 2023-08-26 20:58:46 发布

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本文介绍了一道简单的编程题目，任务是计算指定天数内温度低于或等于35度的日子数量，以便角色RXD可以安排约会。文章提供了完整的代码实现，并提到了题目背景源于对高温天气的吐槽。

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RXD's date

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 1426 Accepted Submission(s): 1085

Problem Description

As we all know that RXD is a life winner, therefore he always goes out, dating with his female friends.
Nevertheless, it is a pity that his female friends don't like extremely hot temperature. Due to this fact, they would not come out if it is higher than 35 degrees.
RXD is omni-potent, so he could precisely predict the temperature in the next

t days, but he is poor in counting.
He wants to know, how many days are there in the next

t days, in which he could go out and date with his female friends.

Input

There is only one test case.

The first line consists of one integer

t.

The second line consists of

t integers

ci which means the temperature in the next

t days.

1≤t≤1000

0≤ci≤50

Output

Output an integer which means the answer.

Sample Input

5

33 34 35 36 37

Sample Output

————————————————————————————————————

签到题，太水了吧，输出比35小的有几个

思路也没什么可说了，出题人都说为了吐槽杭州的高温

真的热了 (.;ﾟ;:_:;ﾟ;.)

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <functional>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;

int n,a[1009];

int main()
{
    while(~scanf("%d",&n))
    {
        int sum=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            if(a[i]<=35) sum++;
        }
        printf("%d\n",sum);
    }
    return 0;
}