HDU-6308 Time Zone(字符串处理)

Problem Description

Chiaki often participates in international competitive programming contests. The time zone becomes a big problem.
Given a time in Beijing time (UTC +8), Chiaki would like to know the time in another time zone s.

Input

There are multiple test cases. The first line of input contains an integer T (1≤T≤106), indicating the number of test cases. For each test case:
The first line contains two integers a, b (0≤a≤23,0≤b≤59) and a string s in the format of "UTC+X'', "UTC-X'', "UTC+X.Y'', or "UTC-X.Y'' (0≤X,X.Y≤14,0≤Y≤9).

Output

For each test, output the time in the format of hh:mm (24-hour clock).

Sample Input

3 11 11 UTC+8 11 12 UTC+9 11 23 UTC+0

Sample Output

11:11 12:12 03:23

题意：

给一个UTC+8时区的时间，求新时区的时间。

思路：

处理字符串将TUC后边的时区数值提取出来，和TUC+8时区做差，

例如UTC-2与UTC+8 相差-10个时区，如果UTC+8时区的时间为12:25，

则UTC-2时区的时间为02:25。

当时不明白负时区如何算，试尽了所有可能，全错了.....

吐槽一下：题目的时区划分不符合地理常识

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int h,m;
char str[20];
int getTime(){//  处理 UTC后的数值，并转化成分钟 
	int flag=0;
	if(str[3]=='+') flag=1;
	else flag=-1;
	int a=0,b=0;
	int i=4,len=strlen(str);
	for(i=4;i<len;i++){
		if(str[i]=='.') break;
		a=a*10+(str[i]-'0');
	}
	if(str[i]=='.') b=str[i+1]-'0';
	int c=a*60+b*6;
	
	return flag*c-8*60; //返回与UTC+8 相差的时间 
}
int main(){
	int T;
	scanf("%d",&T);
	while(T--){
		scanf("%d%d%s",&h,&m,str);
		int t=getTime();
		t=(h*60+m+60*24+t)%(60*24); 
		printf("%02d:%02d\n",t/60,t%60);
	}
	return 0;
}