Time Zone 【模拟时区转换】（HDU暑假2018多校第一场）

传送门：http://acm.hdu.edu.cn/showproblem.php?pid=6308

Time Zone

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5017    Accepted Submission(s): 1433

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

2018 Multi-University Training Contest 1

题意概括：

给出北京时间（北京时区为UTC+8），和需要转换成的目的时间的时区；

求出目的时区的时间；

解题思路：

模拟：

小时和分钟分开处理计算

+时区和 -时区要注意两个加减时间不同的问题，+时区比-时区时间早；

字符转换为整型 atoi(str)

字符转换为双精度浮点型 atof(str)

(当然也可以选择自己手敲一个转换表达式，位数不多）

AC code：

 #include <queue>
 #include <cstdio>
 #include <vector>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 #define INF 0x3f3f3f3f
 #define LL long long
 using namespace std;
 const int MOD1 = ;
 const int MOD2 = ;
 const int MAXN = ;
 char str[MAXN];

 int main()
 {
     int a, b;
     int T_case;
     scanf("%d", &T_case);
     while(T_case--){
         scanf("%d %d %s", &a, &b, &str);
         //printf("%d %d %s\n", a, b, str);
         bool flag = false;
         char aa[MAXN], top1=, bb[MAXN], top2=;
             bb[] = '';
             bb[] = '.';
         for(int i = ; i < strlen(str); i++){

             if(str[i] == '.'){ flag = true; continue;}
             if(!flag){
                 aa[top1++] = str[i];                ///zhengshuu
             }
             else{
                 bb[top2++] = str[i];                ///xiaoshu
             }
         }
         aa[top1] = '\0';bb[top2]='\0';
         double num_it = ;
         int num_b;
         int num_a = atoi(aa);
         if(flag){
                 num_it = atof(bb);
                 num_b = (int)(num_it*);
         }
         //printf("num_a:%d num_b:%d\n", num_a, num_b);
         ///*
         if(str[] == '+'){
             if(num_a >= ) num_a-=;
             else if(num_a < ) num_a = -+num_a;
             if(flag){
                 num_a = num_a+(b+num_b)/;
                 b = (b+num_b)%MOD2;
             }
             a=((a+num_a)%MOD1 + MOD1)%MOD1;
         }

         if(str[] == '-'){

             if(flag){
                 //num_a = num_a-(b+num_b)/60;
                 b = (b-num_b)%MOD2;
                 if(b < ) num_a++;
                 b = (b+MOD2)%MOD2;
             }
             a = ((a-num_a+)%MOD1+MOD1)%MOD1;
         }

         if(a<) printf("0%d:", a);
         else printf("%d:", a);

         if(b<) printf("0%d\n", b);
         else printf("%d\n", b);
         //*/
     }
     return ;
 }