《C语言及程序设计》实践参考——当年第几天
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【项目5-当年第几天】
定义一个函数，其参数为年、月、日的值，返回这一天为该年的第几天。要求在main函数中输入年月日，然后调用这个函数求值，并在main函数中输出结果。
#include<stdio.h>
int days(int y, int m, int d);
int main()
{
int year, month, day;
printf("输入年 月 日: ");
scanf("%d %d %d", &year, &month, &day);
printf("这是该年的第 %d 天\n", days(year, month, day));
return 0;
}
int days(int y, int m, int d)
{ }
[参考解答]
解法1：
#include<stdio.h>
int days(int y, int m, int d);
int main()
{
int year, month, day;
printf("输入年 月 日: ");
scanf("%d %d %d", &year, &month, &day);
printf("这是该年的第 %d 天\n", days(year, month, day));
return 0;
}
int days(int y, int m, int d)
{
int sum=d;
//下面要加上前m-1月的天数
int i;
for(i=1; i<m; i++)
{
switch(i)
{
case 2:
sum+=((y%4==0&&y%100!=0)||y%400==0)?29:28;
break;
case 4:
case 6:
case 9:
case 11:
sum+=30;
break;
default:
sum+=31;
break;
}
}
return sum;
}
解法2：
#include<stdio.h>
int days(int y, int m, int d);
int main()
{
int year, month, day;
printf("输入年 月 日: ");
scanf("%d %d %d", &year, &month, &day);
printf("这是该年的第 %d 天\n", days(year, month, day));
return 0;
}
int days(int y, int m, int d)
{
int sum=d;
//加上前m-1月的天数
int i;
for(i=1; i<m; i++)
{
if(i==1||i==3||i==5||i==7||i==8||i==10||i==12)
sum+=31;
else if (i==4||i==6||i==9||i==11)
sum+=30;
else
sum+=((y%4==0&&y%100!=0)||y%400==0)?29:28;
}
return sum;
}
解法3：后面要学习数组。然后就可以这样来了，30行之内解决问题（对数组充满期待吧）：
#include<stdio.h>
int days(int y, int m, int d);
int main()
{
int year, month, day;
printf("输入年 月 日: ");
scanf("%d %d %d", &year, &month, &day);
printf("这是该年的第 %d 天\n", days(year, month, day));
return 0;
}
int days(int y, int m, int d)
{
int sum=d;
int a[13]= {0,31,28,31,30,31,30,31,31,30,31,30,31};
int i;
for(i=1; i<m; i++)
{
sum+=a[i];
}
if(m>2&&((y%4==0&&y%100!=0)||y%400==0)) //若闰年，且晚于2月，加一天
sum++;
return sum;
}
解法4：有同学写成下面的代码，结果对，但这样的程序的确不好：
#include<stdio.h>
int days(int y, int m, int d);
int main()
{
int year, month, day;
printf("输入年 月 日: ");
scanf("%d %d %d", &year, &month, &day);
printf("这是该年的第 %d 天\n", days(year, month, day));
return 0;
}
int days(int year, int month, int day)
{
int t;
if(year%4==0&&year%100!=0||year%400==0)
{
switch(month)
{
case 1:
t=day;
break;
case 2:
t=day+31;
break;
case 3:
t=day+31+29;
break;
case 4:
t=day+31+29+31;
break;
case 5:
t=day+31+29+31+30;
break;
case 6:
t=day+31+29+31+30+31;
break;
case 7:
t=day+31+29+31+30+31+30;
break;
case 8:
t=day+31+29+31+30+31+30+31;
break;
case 9:
t=day+31+29+31+30+31+30+31+31;
break;
case 10:
t=day+31+29+31+30+31+30+31+31+30;
break;
case 11:
t=day+31+29+31+30+31+30+31+31+30+31;
break;
case 12:
t=day+31+29+31+30+31+30+31+31+30+31+30;
break;
}
return t;
}
else
{
switch(month)
{
case 1:
t=day;
break;
case 2:
t=day+31;
break;
case 3:
t=day+31+28;
break;
case 4:
t=day+31+28+31;
break;
case 5:
t=day+31+28+31+30;
break;
case 6:
t=day+31+28+31+30+31;
break;
case 7:
t=day+31+28+31+30+31+30;
break;
case 8:
t=day+31+28+31+30+31+30+31;
break;
case 9:
t=day+31+28+31+30+31+30+31+31;
break;
case 10:
t=day+31+28+31+30+31+30+31+31+30;
break;
case 11:
t=day+31+28+31+30+31+30+31+31+30+31;
break;
case 12:
t=day+31+28+31+30+31+30+31+31+30+31+30;
break;
}
return t;
}
}