day_of_year另解

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我是靠谱客的博主光亮水蜜桃，最近开发中收集的这篇文章主要介绍day_of_year另解，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

输入年月日，求出是一年中的第多少天，区分闰年和非闰年。

[tiger506@localhost C_C++]$ cat -n doy.c
     1    #include<stdio.h>
     2
     3    int day_of_year(int month, int day, int year)
     4    {
     5    int sum = 0;
     6    switch(month){
     7        case 1: sum = 0 + day;
     8                break;
     9        case 2: sum = 31 + day;
    10                break;
    11        case 3: sum = 31 + 28 + day;
    12                break;
    13        case 4: sum = 31 + 28 + 31 + day;
    14                break;
    15        case 5: sum = 31 + 28 + 31 + 30 + day;
    16                break;
    17        case 6: sum = 31 + 28 + 31 + 30 + 31 + day;
    18                break;
    19        case 7: sum = 31 + 28 + 31 + 30 + 31 + 30 + day;
    20                break;
    21        case 8: sum = 31 + 28 + 31 + 30 + 31 + 30 + 31 + day;
    22                break;
    23        case 9: sum = 31 + 28 + 31 + 30 + 31 + 30 + 31 + 31 + day;
    24                break;
    25        case 10: sum = 31 + 28 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + day;
    26                break;
    27        case 11: sum = 31 + 28 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + 31 + day;
    28                break;
    29        case 12: sum = 31 + 28 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + 31 + 30 + day;
    30                break;
    31        default: break;
    32    }
    33
    34        if(year % 4 == 0 && year % 100 != 0)
    35           sum = sum + 1;
    36        if(year % 400 == 0)
    37           sum = sum + 1;
    38        return sum;
    39    }
    40
    41
    42    int main(void)
    43    {
    44    int month, day, year;
    45    printf("Enter month/day/year :");
    46    scanf("%d/%d/%d", &month, &day, &year);
    47    printf("The sum of day is : %dn", day_of_year(month, day, year));
    48
    49    return 0;
    50    }