hdu6308

Time Zone

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 282    Accepted Submission(s): 94

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

 解析：都换算成分钟，就比较容易算了

#include<bits/stdc++.h>
using namespace std;

#define e exp(1)
#define pi acos(-1)
#define mod 1000000007
#define inf 0x3f3f3f3f
#define ull unsigned long long
#define ll long long

int main()
{
    int t;scanf("%d",&t);
    while(t--)
    {
        char s[40];
        int x,y;scanf( "%d%d%s" ,&x,&y,s);
        int a=0,b=0;
        int len=strlen(s);
        for(int i=4;i<len;i++)
        {
            if(s[i]=='.')break;
            a=a*10+s[i]-'0';
        }
        int f=1;
        for(int i=0;i<len;i++)
            if( s[i]=='.') f=0;
        if( f==0 ) b=s[len-1]-'0';
        int t1=x*60+y,ans;
        if( s[3]=='+'&&a>=8)
            ans=t1+(a-8)*60+b*6;
        else if( s[3]=='+'&&a<8 )
            ans=t1-(8-a)*60+b*6;
        else
            ans=t1-(8+a)*60-b*6;
        ans=(ans+24*60)%(24*60);
        printf("%02d:%02d\n",ans/60,ans%60);
    }
    return 0;
}