hdu 6308 Time Zone （模拟+字符串处理）

Time Zone

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4337    Accepted Submission(s): 1278

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

 

 

题目大意：

给你UTC+8的时间，求给定时区的时间。

 

模拟题。

是水题没错的，但是不仔细做还是会wa得很惨QAQ...

1、这里的时区并不是标准的时区定义，所以直接根据差值加减就好啦。

2、转化成分钟加加减减或许是最好的方法。

3、X可能是两位数。

我是先把给的时间转移到UTC+0(UTC-0)，这样处理起来感觉更方便。

 

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#include<cmath>

using namespace std;

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int a,b;
        char s[10];
        scanf("%d%d%s",&a,&b,s);
        int minute=a*60+b-8*60;

        int add=0,pos;
        for(pos=4;s[pos]!='';pos++)
        {
            if(s[pos]=='.')break;
            add=add*10+s[pos]-'0';
        }
        if(s[pos]=='.')add=add*60+(s[pos+1]-'0')*6;
        else add=add*60;

        if(s[3]=='+')
            minute+=add;
        else
            minute-=add;
        if(minute>=24*60)
            minute=minute-24*60;
        if(minute<0)
            minute=minute+24*60;

        printf("%02d:%02d
",minute/60,minute%60);
    }
    return 0;
}