Question

好的..那么问题就是..假设它是2001年1月1日的星期一所以......作为参考......一年的键盘输入，找出1号的那一天是什么日子那个（输入）年的1月。

在我的程序中（下面）..当人们输入大于参考年份（2001）的输入时..答案是正确的但是..如果输入小于2001 ..答案是错误的。。能否指出并解释我的代码中的错误.. 感谢..

#include <stdio.h>
#include<math.h>
#include <stdlib.h>

int main()
{
    int present_year;
    int normal_days;
    int normal_year;
    int leap_year;
    int leap_days;
    int check;
    int total_days;
    int reference_year=2001;
    int day;

    printf("Enter the year you want to check\n");
    scanf("%d",&present_year);
    if(reference_year<present_year)

    /*if year entered is greater than reference year(2001)*/
    {
        check=present_year-reference_year;
    }

    if(present_year<reference_year)
    /* if year entered is smaller than reference year*/
    {
        check=reference_year-present_year;
    }

    leap_year=check/4;
    normal_year=check-leap_year;
    normal_days=normal_year*365;
    leap_days=leap_year*366;
    total_days=leap_days+normal_days;
    day=total_days%7;

    if(day==0)
        printf("January 1 of year %d will be Monday\n",present_year);
    if(day==1)
        printf("January 1 of year %d will be Tuesday\n",present_year);
    if(day==2)
        printf("January 1 of year %d will be Wednesday\n",present_year);
    if(day==3)
        printf("January 1 of year %d will be Thursday\n",present_year);
    if(day==4)
        printf("January 1 of year %d will be Friday\n",present_year);
    if(day==5)
        printf("January 1 of year %d will be Saturday\n",present_year);
    if(day==6)
        printf("January 1 of year %d will be Sunday\n",present_year);

    return 0;
}

Answer 1

从哪里开始...你甚至了解闰年是什么？我不认为你这样做。它每隔四年不。

在这里，试试这个。

#include <stdio.h>
#include <stdbool.h>
#include <math.h>

double mod(double x, double y)
{
    return x - y * floor(x / y);
}

bool gregorian_leap_year(int year)
{
    return (
        mod(year, 4) == 0 &&
        !(mod(year, 400) == 100 ||
          mod(year, 400) == 200 ||
          mod(year, 400) == 300)) ? true : false;
}

int fixed_from_gregorian(int year, int month, int day)
{
    int correction, f;
    if (month <= 2) correction = 0;
    else if (month > 2 && gregorian_leap_year(year)) correction = -1;
    else correction = -2;

    f = 365 * (year - 1) +
        floor((year - 1) / 4.0) -
        floor((year - 1) / 100.0) +
        floor((year - 1) / 400.0) +
        floor((367 * month - 362) / 12.0) +
        correction + day;
    return f;
}

char *daynames[] = {
    "Sunday",
    "Monday",
    "Tuesday",
    "Wednesday",
    "Thursday",
    "Friday",
    "Saturday"
};

int main(int argc, char *argv[])
{
    int present_year;
    printf("Enter the year you want to check\n");
    scanf("%d", &present_year);

    int f = fixed_from_gregorian(present_year, 1, 1);
    int day = (int)mod(f,7);

    printf("January 1 of year %d will be %s\n", present_year, daynames[day]);

    return 0;
}

Answer 2

我不知道为什么你在2001年1月1日进行了初始化，而你可以在1月1日（即星期一）初始化它。然后只需找到奇数天并计算1月1日XXXX的日期

odd_days=(year-1) + leap_years;

此处年是用户输入

然后

day=odd_days%7;

代码中的错误是什么。？

2 个答案: