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[];
        scanf("%d%d%s",&a,&b,s);
        int minute=a*+b-*;

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

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

        printf("%02d:%02d\n",minute/,minute%);
    }
    return ;
}