【C语言】练习5-8

时间:2014-07-07 19:11:44   收藏:0   阅读:222

  代码实现

#include<stdio.h>

int daytab[2][13] = {
                    {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31},  // 平年的月份
                    {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31},  // 闰年的月份
                    };
/* 把某月某日这种日期表示形式转换为某年中第几天 */
int day_of_year(int year, int month, int day){
    int result = 0;
    int flag = year%4 == 0 && year%100 != 0 || year%400==0; //判断是否为闰年。是则返回1,否则返回0
    int i = 0;
    // 错误检查
    if (year < 1752 || month < 1 || month > 12 || day < 1)
        return -1;
    if(day > daytab[flag][month]){
        return -1;
    }
    // 循环
    for(; i < month; i++){
        result += daytab[flag][i];
    }
    result += day;
    return result;
}

/* 把某年中第几天表示形式转换为某月某日这种日期形式 */
int month_day(int year, int theDays, int *pmonth, int *pday){
    int flag = year%4 == 0 && year%100 != 0 || year%400==0; //判断是否为闰年。是则返回1,否则返回0
    int i = 0;
    // 错误检查
    if (year < 1752 || theDays < 1){
        return -1;
    }
    if ((flag && theDays > 366) || (!flag && theDays > 365)){
        return -1;
    }
    for(; theDays > daytab[flag][i] && i <= 12 ; i++){
        theDays -= daytab[flag][i];
    }
    *pmonth = i;
    *pday = theDays;
    return 0;
}

/* 主函数 */
void main(){
    int month, day;
    printf("%d\n",day_of_year(1988, 3, 1));
    month_day(1988, 61, &month, &day);
    printf("month = %d, day = %d\n", month, day);
}

 

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