Time Zone
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1847 Accepted Submission(s): 560
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
题解:
本来很简单的一道题,因为奇怪的精度损失,导致一直WA,以后注意主要是小数相乘就加上0.1,防止精度损失,因为不知道啥时候有精度损失,所以就全加上
代码:
#include<cstdio>
#include<iostream>
#include<math.h>
using namespace std;
int main()
{
int T;
scanf("%d",&T);
int hh,mm,mmm;
double n;
char a,b,c,d;
while(T--&&scanf("%d %d %c%c%c%lf",&hh,&mm,&a,&b,&c,&n))
{
if(n>=0)
{
if((int)n==n)
{
hh+=24;
hh+=(int)n-8;
hh%=24;
}
else
{
hh+=((int)n-8);
hh+=24;
hh%=24;
double o=60.0*n+0.1;
int k=(int)o;
mmm=k%60;
mm+=mmm;
if(mm>=60)
{
hh++;
hh%=24;
mm%=60;
}
}
}
else if(n<0)
{
n=fabs(n);
if(((int)n==n))
{
hh-=(8+n);
hh+=24;
hh%=24;
}
else
{
hh-=((int)(n+8));
hh+=24;
hh%=24;
double o=60.0*n+0.1;
int k=(int)o;
mmm=k%60;
mm-=mmm;
if(mm<0)
{
hh--;
hh+=24;
hh%=24;
mm+=60;
mm%=60;
}
}
}
if(hh<10) printf("0%d:",hh);
else printf("%d:",hh);
if(mm<10) printf("0%d\n",mm);
else printf("%d\n",mm);
}
}
小数精度,模拟题