【2018多校第一场】hdu6308-Time Zone(日期)
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.
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).
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UTC+8的小时a和分钟b,求 "UTC+X'', "UTC-X'', "UTC+X.Y'', or "UTC-X.Y''的日期是多少。
//转化成分钟计算 #include <bits/stdc++.h> using namespace std; typedef long long LL; int main() { int t, h, m; char s[10]; cin>>t; while(t--) { scanf("%d%d%s", &h, &m, s); h = h*60+m; int op = s[3]=='+'?1:-1; double x; sscanf(s+4, "%lf", &x); x = (int)(x*10+0.005); //读浮点数会有误差 int cha = x*6*op-8*60; h = (h+cha)%(24*60); if(h < 0) h += 24*60; printf("%02d:%02d\n", h/60, h%60); } return 0; }