Time Zone

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 997    Accepted Submission(s): 328

 

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

Source

題目大意：給你時間，讓你求出在經過多少個時區後的時間，

思路：沒有思路，就是直接算

程式碼：

#include<map>
#include<stack>
#include<queue>
#include<cstdio>
#include<string>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
using namespace std;
const int N=4e6+10;
char time[10];
int a[N],b[N];
int main()
{
    int test;
    scanf("%d",&test);
    memset(a,0,sizeof(a));
    memset(b,0,sizeof(b));
    while(test--)
    {
        int i,j,l,k=0,x=0;
        int hour,minute;
        double y=0.0;
        scanf("%d%d",&hour,&minute);
        scanf("%s",time);
        l=strlen(time);
        for(i=4; i<l; i++)
        {
            if(time[i]=='.')
            {
                k=i;
                break;
            }
            else
                x=x*10+(time[i]-'0');
        }
        if(k!=0)
        {
            for(i=k+1; i<l; i++)
            {
                y=0.1*y+(time[i]-'0');
            }
        }
        y*=0.1;
        y*=60;
        if(time[3]=='+')
        {
            x-=8;
            hour+=x;
            minute+=y;
            if(minute>=60)
            {
                minute-=60;
                hour+=1;
            }
            if(hour<0)
            {
                hour+=24;
            }
        }
        else if(time[3]=='-')
        {
            x+=8;
            hour-=x;
            minute-=(int)y;
            if(minute<0)
            {
                hour-=1;
                minute+=60;
            }
            if(minute>=60)
            {
                minute-=60;
                hour+=1;
            }
            if(hour<0)
            {
                hour+=24;
            }
        }
        if(hour>=24)
        {
            hour-=24;
        }
        hour>9?printf("%d:",hour):printf("0%d:",hour);
        minute>9?printf("%d\n",minute):printf("0%d\n",minute);
    }

    return 0;
}