题目:
Friday the ThirteenthIs 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:
Do not use any built-in date functions in your computer language.
Don't just precompute the answers, either, please.
20
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: yang4521 LANG: JAVA TASK: 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()); int count[]=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); } }
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); }