Friday the Thirteenth
Is Friday the 13th really an unusual event?
That is, does the 13th of the month land on a Friday less often than on any other day of the week? To answer this question, write a program that will compute the frequency that the 13th of each month lands on Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, and Saturday over a given period of N years. The time period to test will be from January 1, 1900 to December 31, 1900+N-1 for a given number of years, N. N is non-negative and will not exceed 400.
There are few facts you need to know before you can solve this problem:
January 1, 1900 was on a Monday.
Thirty days has September, April, June, and November, all the rest have 31 except for February which has 28 except in leap years when it has 29.
Every year evenly divisible by 4 is a leap year (1992 = 4*498 so 1992 will be a leap year, but the year 1990 is not a leap year)
The rule above does not hold for century years. Century years divisible by 400 are leap years, all other are not. Thus, the century years 1700, 1800, 1900 and 2100 are not leap years, but 2000 is a leap year.
Do not use any built-in date functions in your computer language.
Don't just precompute the answers, either, please.
PROGRAM NAME: friday
INPUT FORMAT
One line with the integer N.
SAMPLE INPUT (file friday.in)
20
OUTPUT FORMAT
Seven space separated integers on one line. These integers represent the number of times the 13th falls on Saturday, Sunday, Monday, Tuesday, ..., Friday.
SAMPLE OUTPUT (file friday.out)
36 33 34 33 35 35 34
分析:
Brute force is a wonderful thing. 400 years is only 4800 months, so it is perfectly practical to just walk along every month of every year, calculating the day of week on which the 13th occurs for each, and incrementing a total counter.
此题因为输出用的是标准输出而贡献了一个WA,唉………………
/*
ID:xxfz014
PROG:friday
LANG:C++
*/
#include <iostream>
#include <fstream>
using namespace std;
int a[7];
bool isLeap(int year)
{//判断是否是润年
if((year%4==0&&year%100!=0)||(year%400==0)) return true;
return false;
}
int getDays(int month,int year)
{//求得第year年第month月一共有几天
if(month==0) return 0;
if(month==9||month==4||month==6||month==11) return 30;
if(isLeap(year))
{
if(month==2) return 29;
else return 31;
}
else
{
if(month==2) return 28;
else return 31;
}
}
int main()
{
ifstream fin("friday.in");
ofstream fout("friday.out");
int n;
fin>>n;
int sum=0;
for(int i=0;i<n;i++)
{
for(int j=1;j<=12;j++)
{
a[(sum%7+13)%7]++;
sum+=getDays(j,i+1900);
}
}
fout<<a[6]<<" "<<a[0];
for(int i=1;i<=5;i++) fout<<" "<<a[i];
fout<<endl;
fin.close();
fout.close();
return 0;
}