我的第三个 USACO Training--Friday the Thirteenth

题目：

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

因为1900年1月1号是星期一，可以算出各个月距离这一天的天数，然后模七。最后这个题目要注意的是最后输出 每个数有空格，最后一个输出要换行。

我的代码

/*ID: yang4521LANG: JAVATASK: friday*/import java.io.*;import java.util.*;class friday {public static boolean isleapyear(int year){if((year%4==0)&&(year%100!=0))        {            return true;        }        else if(year%400==0)        {            return true;        }else        {            return false;        }}public static void main(String[] args) throws IOException {    BufferedReader f = new BufferedReader(new FileReader("friday.in")); // input file name goes above                                                PrintWriter out = new PrintWriter(new BufferedWriter(new FileWriter("friday.out")));    int N = Integer.parseInt(f.readLine());    intcount[]=new int[7];//These array represent the number of times the 13th falls on  Sunday, Monday, Tuesday, ..., Friday,Saturday    int days=0;//days of one month    int currentYear=1900;    int m;    int allDays=0;    boolean isLeapYear=false;    for(int i=0;i<N;i++)    {        isLeapYear=isleapyear(currentYear);     for(int j=1;j<=12;j++)     {          switch(j)     {     case 4:     case 6:     case 9:     case 11:     days=30;     break;     case 2:     if(isLeapYear)     {     days=29;     }     else     days=28;     break;     default:     days=31;           }    // int a=allDays+1;     count[(allDays+13)%7]++;     allDays+=days;     }             currentYear++;    }    out.print(count[6]+" ");    for(int i=0;i<5;i++)    out.print(count[i]+" "); out.println(count[5]);    out.close();  // close the output file    System.exit(0);  }}

官网：

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>intisleap(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 */intmlen(int y, int m){    if(m == 1)    /* february */        return mtab[m]+isleap(y);    else        return mtab[m];}voidmain(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);}