HDU 5705(思路题）

Clock

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 517    Accepted Submission(s): 180

Problem Description

Given a time HH:MM:SS and one parameter a, you need to calculate next time satisfying following conditions:

1. The angle formed by the hour hand and the minute hand is a.
2. The time may not be a integer(e.g. 12:34:56.78), rounded down(the previous example 12:34:56).

 

Input

The input contains multiple test cases.

Each test case contains two lines.
The first line is the time HH:MM:SS(0≤HH<12,0≤MM<60,0≤SS<60).
The second line contains one integer a(0≤a≤180).

 

Output

For each test case, output a single line contains test case number and the answer HH:MM:SS.

 

Sample Input

0:59:593001:00:0030

 

Sample Output

Case #1: 01:00:00Case #2: 01:10:54

 

Source

"巴卡斯杯" 中国大学生程序设计竞赛 - 女生专场

 

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首先知道时针120秒走1度，分针10秒走一度。故120秒分针和时针相差11度，所以相差一度需要120/11秒，考虑到精度可以让所有的数据乘以11变成整数。时间从120秒开始枚举，直到满足条件结束， 由于前面乘以了11，结果需要除以11.

#include<cstdio>#include<cmath>int main(){    int hh, mm, ss, kase = 0;    while(scanf("%d:%d:%d", &hh, &mm, &ss) == 3){        int a;        scanf("%d", &a);        int t = hh * 3600 + mm * 60 + ss;        t *= 11;        a *= 11;        int i;        int tt = 0, ttt = 0;        for(i = 120; ; i += 120){            tt += 11;            if(tt > 360 * 11) ttt = tt % (11 * 360);            else ttt = tt;            if(ttt > 180 * 11) ttt = 360 * 11 - ttt;            if(ttt == a && i > t) break;        }        i  %= 43200 * 11;        int h = i / (3600 * 11 );        int m = (i - h * (3600 * 11)) / (60 * 11);        int s = (i - h * (3600 * 11) - m * (60 * 11)) / 11;        if(h == 24) h = 0;        printf("Case #%d: %02d:%02d:%02d\n", ++kase, h, m, s);    }}