作者热门文章
- html - 出于某种原因,IE8 对我的 Sass 文件中继承的 html5 CSS 不友好?
- JMeter 在响应断言中使用 span 标签的问题
- html - 在 :hover and :active? 上具有不同效果的 CSS 动画
- html - 相对于居中的 html 内容固定的 CSS 重复背景?
我是 C 语言新手,我编写了这个 C 程序,让用户输入一年中的某一天,作为返回,程序将输出月份以及该月的哪一天。该程序运行良好,但我现在想简化该程序。我知道我需要一个循环,但我不知道如何去做。这是程序
#include <stdio.h>
void SplitDate(int dayofyear, int year, int *month, int *day);
int main() {
int month[13] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int year, dayofyear, *day;
printf("Enter the day of the year: ");
scanf("%d", &dayofyear);
printf("Enter the year: ");
scanf("%d", &year);
printf("Day %d of year %d falls on:\n ",dayofyear, year);
SplitDate(dayofyear, year, month, day);
}
void SplitDate(int dayofyear, int year, int *month, int *day)
{
if(dayofyear >=1 && dayofyear <= 31)
{
printf("month = 1 day = %d\n",dayofyear);
}
else if(dayofyear >=32 && dayofyear <= 59)
{
printf("month = 2 day = %d\n", dayofyear - 31);
}
else if(dayofyear >=60 && dayofyear <=90)
{
printf("month = 3 day = %d\n", dayofyear - 59);
}
else if(dayofyear >=91 && dayofyear <=120)
{
printf("month = 4 day = %d\n", dayofyear - 90);
}
else if(dayofyear >=121 && dayofyear <=151)
{
printf("month = 5 day = %d\n", dayofyear - 120);
}
else if(dayofyear >=151 && dayofyear <=180)
{
printf("month = 6 day = %d\n", dayofyear - 150);
}
else if(dayofyear >=181 && dayofyear <=211)
{
printf("month = 7 day = %d\n", dayofyear - 180);
}
else if(dayofyear >=212 && dayofyear <=242)
{
printf("month = 8 day = %d\n", dayofyear - 211);
}
else if(dayofyear >=243 && dayofyear <=272)
{
printf("month = 9 day = %d\n", dayofyear - 242);
}
else if(dayofyear >=273 && dayofyear <=303)
{
printf("month = 10 day = %d\n", dayofyear -272 );
}
else if(dayofyear >=304 && dayofyear <=333)
{
printf("month = 11 day = %d\n", dayofyear - 303);
}
else if(dayofyear >=334 && dayofyear <=364)
{
printf("month = 12 day = %d\n", dayofyear - 333);
}
}
最佳答案
void SplitDate(int dayofyear, int year, int *month, int *day){
static int months[13] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
//months[2] : Requires a correction of leap year
int i;
for(i=1;i<=12;++i){
if(dayofyear > months[i]){
dayofyear -= months[i];
} else {
printf("month = %d\t day = %d\n", i, dayofyear);
break;
}
}
}
关于C 程序 我想知道是否有办法简化我的 dayofyear 程序?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/23159835/
我是一名优秀的程序员,十分优秀!