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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