Problem J. Bye Bye Russia
Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/gym/100500/attachments

Description

It was the last day at Russia, after the World Finals has ended. Coach fegla was travelling back to Cairo on the same day but later. Coach Fegla was really tired, and he had to sleep before he headed to the Airport. Due to an unknown reason his phone alarm was not working, and he needed something to wake him up. Fortunately his stopwatch was just working fine, and it worked in count down mode, and it supported only minutes. You will be given the current time, and the time he wished to wake up, help him determine the number of needed minutes to configure the stopwatch properly.

Input

The first line will be the number of test cases T. The following T lines each will contain 4 integers hc mc hw mw which represent the current time(hours, and minutes), and the wake up time(hours, minutes). 1 ≤ T ≤ 100 0 ≤ hc, hw ≤ 23 0 ≤ mc, mw ≤ 59 Note that both times will be on the same day, and the wake up time comes after the current time. Assume the day starts at 00:00 and ends at 23:59.

Output

For each test case print a single line containing: Case_x:_y x is the case number starting from 1. y is the required answer. Replace underscores with spaces.

Sample Input

2 1 22 13 21 2 35 21 35

Sample Output

Case 1: 719 Case 2: 1140

HINT

题意

给你两个时间,问你差了多少分钟

题解

水题……

代码

#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define test freopen("test.txt","r",stdin)
const int maxn=;
#define mod 1000000007
#define eps 1e-9
const int inf=0x3f3f3f3f;
const ll infll = 0x3f3f3f3f3f3f3f3fLL;
inline ll read()
{
ll x=,f=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
}
//************************************************************************************* int main()
{
int t=read();
for(int cas=;cas<=t;cas++)
{
int a=read(),b=read(),c=read(),d=read();;
b=a*+b,d=c*+d;
printf("Case %d: %d\n",cas,d-b);
}
}

codeforces Gym 100500 J. Bye Bye Russia的更多相关文章

  1. Codeforces gym 101343 J.Husam and the Broken Present 2【状压dp】

     2017 JUST Programming Contest 2.0 题目链接:Codeforces gym 101343 J.Husam and the Broken Present 2 J. Hu ...

  2. Codeforces GYM 100876 J - Buying roads 题解

    Codeforces GYM 100876 J - Buying roads 题解 才不是因为有了图床来测试一下呢,哼( 题意 给你\(N\)个点,\(M\)条带权边的无向图,选出\(K\)条边,使得 ...

  3. codeforces Gym 100187J J. Deck Shuffling dfs

    J. Deck Shuffling Time Limit: 2   Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100187/pro ...

  4. codeforces GYM 100114 J. Computer Network 无相图缩点+树的直径

    题目链接: http://codeforces.com/gym/100114 Description The computer network of “Plunder & Flee Inc.” ...

  5. codeforces GYM 100114 J. Computer Network tarjan 树的直径 缩点

    J. Computer Network Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100114 Des ...

  6. codeforces gym 100947 J. Killing everything dp+二分

    J. Killing everything time limit per test 4 seconds memory limit per test 64 megabytes input standar ...

  7. codeforces gym 100357 J (网络流)

    题目大意 有n种物品,m种建筑,p个人. n,m,p∈[1,20] 每种建筑需要若干个若干种物品来建造.每个人打算建造一种建筑,拥有一些物品. 主角需要通过交易来建造自己的建筑,交易的前提是对方用多余 ...

  8. Codeforces Gym 100114 J. Computer Network

    Description 给出一个图,求添加一条边使得添加后的图的桥(割边)最少. Sol Tarjan. 一遍Tarjan求割边. 我们发现连接的两个点一定是这两个点之间的路径上的桥最多,然后就可以贪 ...

  9. 【codeforces.com/gym/100240 J】

    http://codeforces.com/gym/100240 J [分析] 这题我搞了好久才搞出样例的11.76....[期望没学好 然后好不容易弄成分数形式.然后我‘+’没打..[于是爆0... ...

随机推荐

  1. Android 生成含签名文件的apk安装包

    做android开发时,必然需要打包生成apk文件,这样才能部署.作为一个完善的apk,必然少不了签名文件,否则下次系统无法进行更新. 一.签名文件的制作及打包生成APK文件 签名文件比较流行的制作方 ...

  2. 头痛的ASCII和preg_replace()

    说这个之前,大家先看下这条语句: preg_replace("/\<\?\=(\\\$[a-zA-Z_\x7f-\xff][a-zA-Z0-9_\[\]\"\'\$\x7f- ...

  3. visual studio 2013 配置 ef+pgsql

    环境:VS2013,WIN7 准备工作: 1.有哪些供应商提供EF6的支持? 可以看msdn给出的答案:Which providers are available for EF6? 在本文使用 Dev ...

  4. 【LeetCode 230】Kth Smallest Element in a BST

    Given a binary search tree, write a function kthSmallest to find the kth smallest element in it. Not ...

  5. 总结:ADO.NET在开发中的部分使用方法和技巧

    如何使用 SqlDataAdapter 来检索多个行 以下代码阐明了如何使用 SqlDataAdapter 对象发出可生成 DataSet 或 DataTable 的命令.它从 SQL Server ...

  6. WeChat Official Account Admin Platform API Introduction

    Keyword: WeChat API Introduction Message and GeneralAuthor: PondBay Studio[WeChat Developer EXPERT] ...

  7. [LeetCode] Ugly Number II (A New Question Added Today)

    Write a program to find the n-th ugly number. Ugly numbers are positive numbers whose prime factors ...

  8. Bias/variance tradeoff

    线性回归中有欠拟合与过拟合,例如下图: 则会形成欠拟合, 则会形成过拟合. 尽管五次多项式会精确的预测训练集中的样本点,但在预测训练集中没有的数据,则不能很好的预测,也就是说有较大的泛化误差,上面的右 ...

  9. Spring学习笔记(一) Spring基础IOC、AOP

    1.       注入类型 a)       Spring_0300_IOC_Injection_Type b)       setter(重要) c)       构造方法(可以忘记) d)     ...

  10. java StreamTokenizer使用

    注意:用JAVA解题一般用Scanner类来进行输入,但对时间要求严格的题,用它可能会超时,我.解POJ1823的时候就遇到这样的问题,后改用StreamTokenizer类进行输入,就过了.看来后者 ...