代码练习系列：问题 B Day of Week

最新推荐文章于 2022-10-26 00:10:38 发布

hanzy88

最新推荐文章于 2022-10-26 00:10:38 发布

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本文介绍了一个程序，用于将特定的俄历日期转换为对应的公历日期，并计算出该日期是星期几。程序首先定义了判断闰年的规则，然后通过一系列函数实现了日期之间的转换与计算。

题目描述

We now use the Gregorian style of dating in Russia. The leap years are years with number divisible by 4 but not divisible by 100, or divisible by 400.
For example, years 2004, 2180 and 2400 are leap. Years 2004, 2181 and 2300 are not leap.
Your task is to write a program which will compute the day of week corresponding to a given date in the nearest past or in the future using today’s agreement about dating.
输入

There is one single line contains the day number d, month name M and year number y(1000≤y≤3000). The month name is the corresponding English name starting from the capital letter.
输出

Output a single line with the English name of the day of week corresponding to the date, starting from the capital letter. All other letters must be in lower case.
样例输入

21 December 2012
5 January 2013
样例输出

Friday
Saturday

#include <stdio.h>
#include <string.h>
#define isleap(x) (x % 100 != 0 && x % 4 == 0)|| x % 400 == 0 ? 1 : 0 

int dayofMonth[13][2] = {{0,0},{31,31},{28,29},{31,31},{30,30},
                        {31,31},{30,30},{31,31},{31,31},{30,30},{31,31},{30,30},{31,31}};

char dayofWeek[7][20] = {"Monday","Tuesday","Wednesday","Thursday","Friday","Saturday","Sunday"};

char monthofYear[13][20] = {"","January","February","March","April","May",
                            "June","July","August","September","October","November","December"};

int cordOfDate(int year1,int month1,int day1){
    int year2 = 2017;
    int month2 = 11;
    int day2 = 8;
    int tempY,tempM,tempD;
    int cnt = 0;
    int num1 = year1*10000 + month1*100 + day1;
    int num2 = year2*10000 + month2*100 + day2;

    if(num1 > num2){
        tempY = year1;
        tempM = month1;
        tempD = day1;
        year1 = year2;
        month1 = month2;
        day1 = day2;
        year2 = tempY;
        month2 = tempM;
        day2 = tempD;
    }

    while(year1 < year2 || month1 < month2 || day1 < day2){
        day1++;
        if(day1 == dayofMonth[month1][isleap(year1)] + 1){
            month1++;
            day1 = 1;
        }

        if(month1 == 13){
            year1++;
            month1 = 1;
        }
        cnt++;
    }

    return cnt;
}


int main()
{
    int day,year;
    char month[20];
    int num,n;

    while(scanf("%d %s %d",&day,month,&year) != EOF){
        for(int i = 0;i < 14;i++){
            if(strcmp(month,monthofYear[i]) == 0){
                num = i;
                break;
            }
        }
        printf("%d\n",num);
        n = cordOfDate(year,num,day);
        printf("%d\n",n);

        puts(dayofWeek[(n+2) % 7]);
    }
    return 0;
}