
给定的日期格式为日期、月份和年份(整数)。任务是确定该日期是否可行。
有效日期范围应为 1/1/1800 – 31/12/9999,超出这些日期的日期无效。
这些日期不仅包含年份范围,还包含与日历日期相关的所有约束。
约束是 -
如果所有约束都为真,则它是有效日期,否则不是。
Input: y = 2002 d = 29 m = 11 Output: Date is valid Input: y = 2001 d = 29 m = 2 Output: Date is not valid
START
In function int isleap(int y)
Step 1-> If (y % 4 == 0) && (y % 100 != 0) && (y % 400 == 0) then,
Return 1
Step 2-> Else
Return 0
In function int datevalid(int d, int m, int y)
Step 1-> If y < min_yr || y > max_yr then,
Return 0
Step 2-> If m < 1 || m > 12 then,
Return 0
Step 3-> If d < 1 || d > 31 then,
Return 0
Step 4-> If m == 2 then,
If isleap(y) then,
If d <= 29 then,
Return 1
Else
Return 0
End if
End if
Step 5-> If m == 4 || m == 6 || m == 9 || m == 11 then,
If(d <= 30)
Return 1
Else
Return 0
Return 1
End Function
In main(int argc, char const *argv[])
Step 1->Assign and initialize values as y = 2002, d = 29, m = 11
Step 2-> If datevalid(d, m, y) then,
Print "Date is valid"
Step 3-> Else
Print "date is not valid”
End main
STOP实时演示
#include <stdio.h>
#define max_yr 9999
#define min_yr 1800
//to check the year is leap or not
//if the year is a leap year return 1
int isleap(int y) {
if((y % 4 == 0) && (y % 100 != 0) && (y % 400 == 0))
return 1;
else
return 0;
}
//Function to check the date is valid or not
int datevalid(int d, int m, int y) {
if(y < min_yr || y > max_yr)
return 0;
if(m < 1 || m > 12)
return 0;
if(d < 1 || d > 31)
return 0;
//Now we will check date according to month
if( m == 2 ) {
if(isleap(y)) {
if(d <= 29)
return 1;
else
return 0;
}
}
//April, June, September and November are with 30 days
if ( m == 4 || m == 6 || m == 9 || m == 11 )
if(d <= 30)
return 1;
else
return 0;
return 1;
}
int main(int argc, char const *argv[]) {
int y = 2002;
int d = 29;
int m = 11;
if(datevalid(d, m, y))
printf("Date is valid</p><p>");
else
printf("date is not valid</p><p>");
return 0;
}如果运行上面的代码,它将生成以下输出 -
Date is valid
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