题目链接
题目描述
The World can indicate world travel, particularly on a large scale. You mau be lucky enough to be embarking on a six-month overseas trip, or are working, studying or living overseas for an extended period of time. Similarly, this card reinforces Universal understanding and globalawareness， and you will have a new appliciation for people and cultures from across the world.
Across the world there are various time zones, leading to time differences. Here you are informed of several famous capitals and their corresponding time zones.
• Beijing - China - UTC + 8 (China Standard Time)
• Washington - United States - UTC - 5 (Eastern Standard Time)
• London United Kingdom - UTC (Greenwich Mean Time)
• Moscow - Russia － UTC + 3 (Moscow Time)
Given t he local time of a city, you are expected to calculate the date and local time of another specific city among the above capitals.
输入
The first line of input contains a single integer T≤1000 indicating the number of testcases.
Each testcase consists of three lines. The first line is in the form of "hour:minute AM/ PM”(1≤hour≤12, 00≤minute≤59) indicating the local time. Next two lines contain two strings s1, s2. s1 is the name of city corresponding to the given time, while s2 indicates the city you are expected to calculate the local time.
输出
For each testcase, begin with “Case i:", where i indicate the case number, and then output a single line in the following format “Yesterday / Today / Tomorrow hour:minute AM/ PM”, separated by spaces. The first word describes the corresponding date.
样例输入
2
12:00 AM
London
Moscow
4:00 PM
London
Beijing样例输出
Case 1: Today 3:00 AM
Case 2: Tomorrow 12:00 AM
题意
已知四个城市之间的时差，已知一个城市的时间，求另一个城市的时间，时间是12进制的，还要分昨天今天明天还有上午下午
题解
转化成24进制的时间又快又不容易出错（之前用12进制WA了N发）
代码
#include
using namespace std;
#define rep(i,a,n) for(int i=a;i
#define PLL pair
#define PI acos(1.0)
#define eps 1e-6
#define inf 1e17
#define INF 0x3f3f3f3f
#define N 205
const int maxn = 1e5+5;
char d[3][100] = {"Yesterday","Today","Tomorrow"};
char city[4] = {'B','W','L','M'};
int ti[4] = {8,-5,0,3};
map p;
int t,hh,mm;
char am[2],c1[100],c2[100],day[100];
int kase = 0;
int main()
{
rep(i,0,4)
p[city[i]] = ti[i];
sca(t);
while(t--)
{
scanf("%d:%d",&hh,&mm);
scs(am);
getchar();
scs(c1);
scs(c2);
if(hh==12 && am[0] =='A')
hh = 0;
if(hh<12&&am[0] =='P')
hh+=12;
hh += p[c2[0]] - p[c1[0]];
if(hh < 0)
{
strcpy(day,d[0]);
hh += 24;
}
else if(hh < 24)
{
strcpy(day,d[1]);
}
else
{
strcpy(day,d[2]);
hh-=24;
}
if(hh==0)
{
hh = 12;
am[0] = 'A';
}
else if(hh > 0 && hh < 12)
{
am[0] = 'A';
}
else if(hh==12)
{
am[0] = 'P';
}
else
{
hh -= 12;
am[0] = 'P';
}
printf("Case %d: %s %d:%02d %s\n",++kase,day,hh,mm,am);
}
}