Time

最新推荐文章于 2022-10-26 18:18:30 发布

Apollo_Dou

最新推荐文章于 2022-10-26 18:18:30 发布

阅读量301

点赞数

CC 4.0 BY-SA版权

分类专栏： ACM

本文链接：https://blog.youkuaiyun.com/Kyrene_D/article/details/68947801

ACM 专栏收录该内容

21 篇文章

订阅专栏

本文介绍了一个简单的程序设计问题，即如何判断两个给定的时间是否相等。通过将时间转换为秒进行比较，解决了一道编程练习题。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

                                   Time

A plane can go from city X to city Y in 1 hour and 20 minutes. However, when it returns from city Y to city X, with the same speed, it spends only 80 minutes! If you don’t know how might that happen, just remember that 1 hour and 20 minutes is the same as 80 minutes ;-) In this problem you will have 2 durations of time and you have to decide if they are equal or not.

Input
The first line will be the number of test cases T. Each test case has two lines, the first line represents the first duration and contains 3 integers: h, m, s; these are the Hours, Minutes and Seconds. The second line represent the second duration with 3 integers also, in the same way. All integers are positive and below 5000.

Output
For each test case, print one line which contains the number of the test case. Then print “Yes” if the 2 durations are equal and print “No” otherwise, see the samples and follow the output format.

Example
Input
3
1 20 0
0 80 0
0 2 10
0 0 130
2 10 5
2 11 0
Output
Case 1: Yes
Case 2: Yes
Case 3: No

题目大意：就是两个时间，判断他们是不是表示同一个时间；

思路：大水题，将小时、分钟都换成秒，最后再比较就行。
//没看题直接把Yes和No都打成大写的了，我好晕啊……

code：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define INF 0x3f3f3f3f
#define LL long long

int h1,m1,s1,h2,m2,s2;

int main()
{
    int t,tt = 0;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d%d",&h1,&m1,&s1);
        scanf("%d%d%d",&h2,&m2,&s2);
        m1 += h1*60;
        m2 += h2*60;
        s1 += m1*60;
        s2 += m2*60;
        //printf("%d %d\n",s1,s2);
        if(s1==s2)
            printf("Case %d: Yes\n",++tt);
        else
            printf("Case %d: No\n",++tt);
    }
    return 0;
}