• HDU-6308 Time Zone(时区转换)


    Time Zone
    Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 2866 Accepted Submission(s): 908

    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:给出北京时间,给出要变化的时区,将输出当地时间。
    将小时直接转换成分钟,注意针对北京+8来调整相差时间的正负,还有处理小数时区,直接将小时转换成分钟后0.1小时即6分钟

    #include<bits/stdc++.h>
    using namespace std;
    int t,a,b,x,y;
    int main()
    {
        char tim[10];
        scanf("%d",&t);
        while(t--)
        {
            scanf("%d%d",&a,&b);
            scanf("%s",tim);
            x=0,y=0;
            for(int i=4;i<strlen(tim);i++)
            {
                if(tim[i]=='.')
                {
                    y=tim[i+1]-'0';
                    break;
                }
                x*=10;
                x+=tim[i]-'0';
            }
            x=x*60+y*6;
            if(tim[3]=='-')x=-x;
            x-=480;
            a=a*60+b;
            a=(a+x+1440)%1440;
            printf("%02d:%02d
    ",a/60,a%60);
        }
    }
    
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  • 原文地址:https://www.cnblogs.com/kuronekonano/p/11135737.html
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