Time Zone
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5204 Accepted Submission(s): 878
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
#include <cstdio>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
int main() {
std::ios::sync_with_stdio(false);
int T;
std::cin >> T;
for (int cas = 1; cas <= T; ++cas) {
int sgn;
int a, b, x, y(0);
std::string s;
std::cin >> a >> b >> s;
sgn = (s[3] == '+' ? 1 : -1);
s = s.substr(4);
/*
假设:string s = "0123456789";
string sub1 = s.substr(5); //只有一个数字5表示从下标为5开始一直到结尾:sub1 = "56789"
string sub2 = s.substr(5, 3); //从下标为5开始截取长度为3位:sub2 = "567"
*/
if (s.size() > 2) {
y = s.back() - '0';
s.pop_back();
s.pop_back();
}
x = std::stoi(s);
int m = (a + 24) * 60 + b;
int c = 80;
int d = sgn * (x * 10 + y);
int e = d - c;
int f = e * 6;
int g = (m + f) % 1440;
printf("%02d:%02d\n", g / 60, g % 60);
}
return 0;
}