题解 | #计算日期到天数转换#
计算日期到天数转换
https://www.nowcoder.com/practice/769d45d455fe40b385ba32f97e7bcded
#include <stdio.h>
//第一种简单数组方法
int main()
{
int year;
int month;
int day;
scanf("%d %d %d", &year, &month, &day);
int sum = day;
//闰年数组：
int leap_year[12] = { 31,29,31,30,31,30,31,31,30,31,30,31 };
// 1 2 3 4 5 6 7 8 9 10 11 12
//平年数组：
int common_year[12] = { 31,28,31,30,31,30,31,31,30,31,30,31 };
// 1 2 3 4 5 6 7 8 9 10 11 12
//闰年情况：
if ((year % 4 == 0 && year % 100 != 0) || (year % 400 == 0))
{
if (month == 1)
{
sum = day;
}
else
{
int j = 0;
for (int i = 1; i < month; i++)
{
sum = sum + leap_year[j++];
}
}
}
//平年情况：
else
{
if (month == 1)
{
sum = day;
}
else
{
int j = 0;
for (int i = 1; i < month; i++)
{
sum = sum + common_year[j++];
}
}
}
printf("%d", sum);
return 0;
}
//第二种暴力方法：
int main()
{
int year;
int month;
int day;
scanf("%d %d %d", &year, &month, &day);
int sum = 0;
//1 3 5 7 8 10 12 ——31天
//4 6 9 11 ——30天
//闰年2月 ——29天
//平年2月 ——28天
//如果year是闰年 2000 11 15
if ((year % 4 == 0 && year % 100 != 0) || (year % 400 == 0))
{
switch (month)
{
case 1:
sum = day;
break;
case 2:
sum = 31 + day;
break;
case 3:
sum = 31 + 29 + day;
break;
case 4:
sum = 31 + 29 + 31 + day;
break;
case 5:
sum = 31 + 29 + 31 + 30 + day;
break;
case 6:
sum = 31 + 29 + 31 + 30 + 31 + day;
break;
case 7:
sum = 31 + 29 + 31 + 30 + 31 + 30 + day;
break;
case 8:
sum = 31 + 29 + 31 + 30 + 31 + 30 + 31 + day;
break;
case 9:
sum = 31 + 29 + 31 + 30 + 31 + 30 + 31 + 31 + day;
break;
case 10:
sum = 31 + 29 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + day;
break;
case 11:
sum = 31 + 29 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + 31 + day;
break;
case 12:
sum = 31 + 29 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + 31 + 30 + day;
break;
}
}
else
{
switch (month)
{
case 1:
sum = day;
break;
case 2:
sum = 31 + day;
break;
case 3:
sum = 31 + 28 + day;
break;
case 4:
sum = 31 + 28 + 31 + day;
break;
case 5:
sum = 31 + 28 + 31 + 30 + day;
break;
case 6:
sum = 31 + 28 + 31 + 30 + 31 + day;
break;
case 7:
sum = 31 + 28 + 31 + 30 + 31 + 30 + day;
break;
case 8:
sum = 31 + 28 + 31 + 30 + 31 + 30 + 31 + day;
break;
case 9:
sum = 31 + 28 + 31 + 30 + 31 + 30 + 31 + 31 + day;
break;
case 10:
sum = 31 + 28 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + day;
break;
case 11:
sum = 31 + 28 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + 31 + day;
break;
case 12:
sum = 31 + 28 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + 31 + 30 + day;
break;
}
}
printf("%d", sum);
return 0;
}
