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

Input

There are multiple test cases. The first line of input contains an integer

Output

For each test, output the time in the format of

Sample Input

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：

 1 #include <queue>
 2 #include <cstdio>
 3 #include <vector>
 4 #include <cstring>
 5 #include <iostream>
 6 #include <algorithm>
 7 #define INF 0x3f3f3f3f
 8 #define LL long long
 9 using namespace std;
10 const int MOD1 = 24;
11 const int MOD2 = 60;
12 const int MAXN = 20;
13 char str[MAXN];
14 
15 int main()
16 {
17     int a, b;
18     int T_case;
19     scanf("%d", &T_case);
20     while(T_case--){
21         scanf("%d %d %s", &a, &b, &str);
22         //printf("%d %d %s
", a, b, str);
23         bool flag = false;
24         char aa[MAXN], top1=0, bb[MAXN], top2=2;
25             bb[0] = '0';
26             bb[1] = '.';
27         for(int i = 4; i < strlen(str); i++){
28 
29             if(str[i] == '.'){ flag = true; continue;}
30             if(!flag){
31                 aa[top1++] = str[i];                ///zhengshuu
32             }
33             else{
34                 bb[top2++] = str[i];                ///xiaoshu
35             }
36         }
37         aa[top1] = '';bb[top2]='';
38         double num_it = 0;
39         int num_b;
40         int num_a = atoi(aa);
41         if(flag){
42                 num_it = atof(bb);
43                 num_b = (int)(num_it*60);
44         }
45         //printf("num_a:%d num_b:%d
", num_a, num_b);
46         ///*
47         if(str[3] == '+'){
48             if(num_a >= 8) num_a-=8;
49             else if(num_a < 8) num_a = -8+num_a;
50             if(flag){
51                 num_a = num_a+(b+num_b)/60;
52                 b = (b+num_b)%MOD2;
53             }
54             a=((a+num_a)%MOD1 + MOD1)%MOD1;
55         }
56 
57         if(str[3] == '-'){
58 
59             if(flag){
60                 //num_a = num_a-(b+num_b)/60;
61                 b = (b-num_b)%MOD2;
62                 if(b < 0) num_a++;
63                 b = (b+MOD2)%MOD2;
64             }
65             a = ((a-num_a+16)%MOD1+MOD1)%MOD1;
66         }
67 
68         if(a<10) printf("0%d:", a);
69         else printf("%d:", a);
70 
71         if(b<10) printf("0%d
", b);
72         else printf("%d
", b);
73         //*/
74     }
75     return 0;
76 }