ZOJ.3131 Digital Clock【打表】 2015/10/11

Digital Clock

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Time Limit: 1 Second      Memory Limit: 32768 KB

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Digital clocks usually show the time in the form hh:mm:ss, where hh is a number between 00 and 23, and both mm and ss are numbers between 00 and 59. Removing the colons from hh:mm:ss will result in an integer hhmmss, which is called aclock integer. For example, the clock integer of 17:05:13 is 170513 and the clock integer of 00:07:37 is 737.

You are given a time interval and you are to determine the number of clock integers in it that are multiples of 3. A time interval will be given by specifying its start and end time. For example, the time interval [00:59:58, 01:01:24] has a total of 1+1+60+25=87 clock integers, namely, 5958, 5959, 10000 through 10059, and 10100 through 10124. How many of them are multiples of 3?

Note that a time interval that includes midnight may have a start time greater than its end time, as in [22:47:03, 01:03:24]. You may assume that a time interval is at least one second long and shorter than 24 hours.

Write a program that can determine the number of multiples of 3 in a time interval.

Input

Your program is to read from standard input. The input consists of T test cases. The number of test casesT is given in the first line of the input. Each test case consists of a single line that contains the start time and end time of a time interval, which are separated by a single space.

Output

Your program is to write to standard output. Each test case outputs exactly one line. Print the number of multiples of 3 among the clock integers in the time interval. The following shows a sample input with three test cases and its output.

Sample Input

300:59:58 01:01:2422:47:03 01:03:2400:00:09 00:03:37

Sample Output

29272770

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Source: Asia Regional Contest Seoul 2006

打表做，记录从0:00:00到当前时刻符合要求的时刻的数量，如果是第一天到第二天，分两次求

#include<iostream>#include<cstdio>#include<cstring>using namespace std;int times[300000];void init(){    memset(times,0,sizeof(times));    int i,j,k;    times[0] = 1;    for( i = 1 ; i <= 235959 ; ++i ){        if( i % 100 >= 60 || i % 10000 >= 6000 )            times[i] = -1;        else{            if( i % 10000 == 0 ){                times[i] = times[i-4041];                if( i % 3 == 0 )                    times[i]++;            }            else if( i % 100 == 0 ){                times[i] = times[i-41];                if( i % 3 == 0 )                    times[i]++;            }            else{                times[i] = times[i-1];                if( i % 3 == 0 )                    times[i]++;            }        }    }}int main(){    int h,h1,h2,m,m1,m2,s,s1,s2;    int t,ans;    int ret1,ret2;    init();    cin>>t;    while(t--){        scanf("%d:%d:%d %d:%d:%d",&h1,&m1,&s1,&h2,&m2,&s2);        ret1 = h1 * 10000 + m1 * 100 + s1;        ret2 = h2 * 10000 + m2 * 100 + s2;        if( ret1 > ret2 ){            ans = times[235959] - times[ret1];            ans += times[ret2];        }        else ans = times[ret2] - times[ret1];        if( ret1 % 3 ==0 )            ans++;        cout<<ans<<endl;    }    return 0;}