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2015
05-23

HDU 4177-Avoiding a disaster-模拟-[解题报告]HOJ

Avoiding a disaster

问题描述 :

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?

输入:

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.

输出:

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.

样例输入:

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

样例输出:

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

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
 

 

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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;
}

 

参考:http://www.cnblogs.com/13224ACMer/p/4728968.html