HDU 4177 模拟时间问题

最新推荐文章于 2018-08-01 19:57:00 发布

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最新推荐文章于 2018-08-01 19:57:00 发布

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文章标签： java

原文链接：http://www.cnblogs.com/13224ACMer/p/4728968.html

本文介绍了解决三只手表显示不同时间且一只显示错误时间的问题，通过编写程序匹配正确的实际时间。提供了输入和输出样例，以及解决逻辑。

Avoiding a disaster

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 211 Accepted Submission(s): 101

Problem Description

Percy likes to be punctual. So much so that he always keeps three watches with him, so that he can be sure exactly what the time is. However, Percy's having a bad day. He found out that one of his watches was giving the wrong time. What's worse, when he went to correct the watch, he corrected the wrong one! That is, one watch was running x minutes behind (where x <= 480) and he wound one of the other watches x minutes forward. He now has three watches reading three different times, and hence is in serious danger of being tardy. Can you help Percy by writing a program that takes in the three times displayed on the watches and returns the correct time?

Input

The input begins with an integer T indicating the number of cases that follow (0 < T
< 100). Each of the following T lines contains one test case, made up of three readings, separated by single space characters: H1:M1 H2:M2 H3:M3 In each reading H1,H2,H3 represent the hours displayed (0 < H1,H2,H3 < 13), and M1,M2,M3 represent the minutes displayed (0 <= M1,M2,M3 < 60).
If the number of minutes is less than 10, a leading 0 is added.

Output

For each test case, one line should be produced, formatted exactly as follows: "The
correct time is Hi:Mi". If the number of minutes is less than 10, a leading 0 should be
added. If the number of hours is less than 10, a leading 0 should NOT be added. If it is impossible to tell the time from the three readings, print the string: "Look at the sun".

Sample Input

   3 5:00 12:00 10:00 11:59 12:30 1:01 12:00 4:00 8:00
  

Sample Output

   The correct time is 5:00 The correct time is 12:30 Look at the sun
  

Source

The 2011 South Pacific Programming Contest

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#include<stdio.h>
#include<string.h>
#include<math.h>
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        char c1,c2,c3;
        int h1,h2,h3;
        char m11,m12,m21,m22,m31,m32;
        scanf("%d%c%c%c",&h1,&c1,&m11,&m12);
        getchar();
        scanf("%d%c%c%c",&h2,&c2,&m21,&m22);
        getchar();
        scanf("%d%c%c%c",&h3,&c3,&m31,&m32);
        int min1=0,min2=0,min3=0;
        min1=h1*60+m11*10+m12;
        min2=h2*60+m21*10+m22;
        min3=h3*60+m31*10+m32;
        int p1=0,p2=0,p3=0;
        if(min1+min2==2*min3&&fabs(min1-min3)<=480)
            p3++;
        else if(min1+min3==2*min2&&fabs(min1-min2)<=480)
            p2++;
        else if(min3+min2==2*min1&&fabs(min3-min1)<=480)
            p1++;
        min1+=12*60;
        if(min1+min2==2*min3&&fabs(min1-min3)<=480)
            p3++;
        else if(min1+min3==2*min2&&fabs(min1-min2)<=480)
            p2++;
        else if(min3+min2==2*min1&&fabs(min3-min1)<=480)
            p1++;
        min1-=12*60;
        min2+=12*60;
        if(min1+min2==2*min3&&fabs(min1-min3)<=480)
            p3++;
        else if(min1+min3==2*min2&&fabs(min1-min2)<=480)
            p2++;
        else if(min3+min2==2*min1&&fabs(min3-min1)<=480)
            p1++;
        min2-=12*60;
        min3+=12*60;
        if(min1+min2==2*min3&&fabs(min1-min3)<=480)
            p3++;
        else if(min1+min3==2*min2&&fabs(min1-min2)<=480)
            p2++;
        else if(min3+min2==2*min1&&fabs(min3-min1)<=480)
            p1++;
        min1+=12*60;
        min2+=12*60;
        min3-=12*60;
        if(min1+min2==2*min3&&fabs(min1-min3)<=480)
            p3++;
        else if(min1+min3==2*min2&&fabs(min1-min2)<=480)
            p2++;
        else if(min3+min2==2*min1&&fabs(min3-min1)<=480)
            p1++;
        min2-=12*60;
        min3+=12*60;
        if(min1+min2==2*min3&&fabs(min1-min3)<=480)
            p3++;
        else if(min1+min3==2*min2&&fabs(min1-min2)<=480)
            p2++;
        else if(min3+min2==2*min1&&fabs(min3-min1)<=480)
            p1++;
        min1-=12*60;
        min2+=12*60;
        if(min1+min2==2*min3&&fabs(min1-min3)<=480)
            p3++;
        else if(min1+min3==2*min2&&fabs(min1-min2)<=480)
            p2++;
        else if(min3+min2==2*min1&&fabs(min3-min1)<=480)
            p1++;
        min1+=12*60;
        if(min1+min2==2*min3&&fabs(min1-min3)<=480)
            p3++;
        else if(min1+min3==2*min2&&fabs(min1-min2)<=480)
            p2++;
        else if(min3+min2==2*min1&&fabs(min3-min1)<=480)
            p1++;
        if(p1>0&&p2>0||p1>0&&p3>0||p2>0&&p3>0)
            printf("Look at the sun\n");
        else if(p1>0)
        {
            printf("The correct time is %d:%c%c\n",h1,m11,m12);
        }
        else if(p2>0)
        {
            printf("The correct time is %d:%c%c\n",h2,m21,m22);
        }
        else if(p3>0)
        {
            printf("The correct time is %d:%c%c\n",h3,m31,m32);
        }
    }
    return 0;
}