year年有366天,而正常年份有365天,任务是通过程序检查给定年份是否为a年。
它的逻辑可以通过检查是否将Year除以400或4,但是如果数字未除以正常年份而得到的任何数字。
Input-: year=2000 Output-: 2000 is a Leap Year Input-: year=101 Output-: 101 is not a Leap year
Start Step 1 -> declare function 布尔检查年份是否为a年 bool check(int year) IF year % 400 = 0 || year%4 = 0 return true End Else return false End Step 2 -> In main() Declare variable as int year = 2000 Set check(year)? printf("%d is a Leap Year",year): printf("%d is not a Leap Year",year) Set year = 10 Set check(year)? printf("%d is a Leap Year",year): printf("\n%d is not a Leap Year",year); Stop
#include <stdio.h> #include <stdbool.h> //布尔检查年份是否为a年 bool check(int year){ //如果一年是400的倍数或4的倍数,则它是a年 if (year % 400 == 0 || year%4 == 0) return true; else return false; } int main(){ int year = 2000; check(year)? printf("%d is a Leap Year",year): printf("%d is not a Leap Year",year); year = 101; check(year)? printf("%d is a Leap Year",year): printf("\n%d is not a Leap Year",year); return 0; }
输出结果
2000 is a Leap Year 101 is not a Leap Year