c Language Leap Year Common Year Problem(1800~2800)

); scanf ( "%d" , &year); // 1. 接收用户输入的年份 if (year < 0 || year > 3000 ) // 2. 检查年份是否在有效范围内 { printf ( "年份超出范围。\n" ); } else { if (year % 4 == 0 ) // 3. 检查能否被4整除 { if (year % 100 == 0 ) // 4. 检查能否被100整除 { if (year % 400 == 0 ) // 5. 检查能否被400整除 { flag = 1 ; } else { flag = 0 ; } } else { flag = 1 ; } } else { flag = 0 ; } if (flag) // 6. 根据flag的值判断是否为闰年 { printf ( "%d 是闰年。\n" , year); } else { printf ( "%d 不是闰年。\n" , year); } } return 0 ; } .

This answer quotes ChatGPT

Full C code implementation, using the if-else structure and scanf function to get the year the user entered:

#include 

int main() {
    int year, flag;
    printf("请输入年份：");
    scanf("%d", &year);
    if (year < 1800 || year > 2800) {
        printf("年份超出范围。");
    } else {
        if (year % 4 == 0) {
            if (year % 100 != 0) {
                flag = 1;
            } else {
                if (year % 400 == 0) {
                    flag = 1;
                } else {
                    flag = 0;
                }
            }
        } else {
            flag = 0;
        }
        if (flag == 1) {
            printf("%d 是闰年。\n", year);
        } else {
            printf("%d 不是闰年。\n", year);
        }
    }
    return 0;
}

1. Obtain the year entered by the user through the scanf function and store it in the year variable.
2. If the entered year is not in the range of 1800 to 2800, an error message is displayed, and the program ends.
3. If the year is divisible by 4, go to Step 4. Otherwise, go to Step 6.
4. If the year is divisible by 100, perform Step 5. Otherwise, it means leap year, flag = 1.
5. If the year is divisible by 400, it is a leap year, marked as flag = 1; Otherwise, it is not a leap year and flag = 0.
6. If the year is not divisible by 4, it is not a leap year and flag = 0.
7. Judge the result according to flag.