Friday the Thirteenth

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 positive and will not exceed 400.

Note that the start year is NINETEEN HUNDRED, not 1990.

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：

	1900年1月1日是星期一所以1899年的最后一天是星期天，计算1900年1月13日只需计算13%7即可。公式：；计算13在星期几（13%7+a）%7；

a：（星期？）上一个月的：a=(a+(31||30||29||28)%7)%7(a的初始化值为0)
代码：
#include <fstream>
using namespace std;
int main()
{
    ofstream fout("friday.out");
    ifstream fin("friday.in");
    int num;
    char month[12]={3,0,3,2,3,2,3,3,2,3,2,3};
    int week[7]={0};
    while(fin>>num){
    for(int i=0,key=7,year=1900,last=0;i<num;i++,year++)
    {
        ((year%4==0&&year%100!=0)||(year%400==0))?month[1]=1:month[1]=0;
        for(int j=0;j<12;j++)
        {
            week[(key+last)%7]++;
            last=(last+month[j])%7;
        }
    }
    for(int i=0;i<7;i++)
    {fout<<week[i];if(i<6)fout<<" ";week[i]=0;}
    fout<<endl;
    }
}
结果：
USER: jim zhai [jszhais1]
TASK: friday
LANG: C++

Compiling...
Compile: OK

Executing...
   Test 1: TEST OK [0.000 secs, 3360 KB]
   Test 2: TEST OK [0.000 secs, 3360 KB]
   Test 3: TEST OK [0.000 secs, 3360 KB]
   Test 4: TEST OK [0.000 secs, 3360 KB]
   Test 5: TEST OK [0.000 secs, 3360 KB]
   Test 6: TEST OK [0.000 secs, 3360 KB]
   Test 7: TEST OK [0.000 secs, 3360 KB]
   Test 8: TEST OK [0.000 secs, 3360 KB]

All tests OK.
Your program ('friday') produced all correct answers!  This is your
submission #5 for this problem.  Congratulations!
Here are the test data inputs:
------- test 1 ----
1
------- test 2 ----
2
------- test 3 ----
5
------- test 4 ----
13
------- test 5 ----
45
------- test 6 ----
100
------- test 7 ----
256
------- test 8 ----
400
Keep up the good work!
Thanks for your submission!
答案：
Friday the Thirteenth
Russ Cox
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.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <assert.h>

int
isleap(int y)
{
    return y%4==0 && (y%100 != 0 || y%400 == 0);
}

int mtab[] = { 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };

/* return length of month m in year y */
int
mlen(int y, int m)
{
    if(m == 1)    /* february */
        return mtab[m]+isleap(y);
    else
        return mtab[m];
}

void
main(void)
{
    FILE *fin, *fout;
    int i, m, dow, n, y;
    int ndow[7];

    fin = fopen("friday.in", "r");
    fout = fopen("friday.out", "w");
    assert(fin != NULL && fout != NULL);

    fscanf(fin, "%d", &n);

    for(i=0; i<7; i++)
        ndow[i] = 0;

    dow = 0;    /* day of week: January 13, 1900 was a Saturday = 0 */
    for(y=1900; y<1900+n; y++) {
        for(m=0; m<12; m++) {
            ndow[dow]++;
            dow = (dow+mlen(y, m)) % 7;
        }
    }

    for(i=0; i<7; i++) {
        if(i)
            fprintf(fout, " ");
        fprintf(fout, "%d", ndow[i]);
    }
    fprintf(fout, "\n");

    exit(0);
}