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【项目5-当年第几天】
定义一个函数，其参数为年、月、日的值，返回这一天为该年的第几天。要求在main函数中输入年月日，然后调用这个函数求值，并在main函数中输出结果。
#include int days(int y, int m, int d);int main(){ int year, month, day; printf("输入年 月 日: "); scanf("%d %d %d", &year, &month, &day); printf("这是该年的第 %d 天\n", days(year, month, day)); return 0;}int days(int y, int m, int d){ }
[参考解答]
解法1：
#include int days(int y, int m, int d);int main(){ int year, month, day; printf("输入年 月 日: "); scanf("%d %d %d", &year, &month, &day); printf("这是该年的第 %d 天\n", days(year, month, day)); return 0;}int days(int y, int m, int d){ int sum=d; //下面要加上前m-1月的天数 int i; for(i=1; i
解法2：
#include int days(int y, int m, int d);int main(){ int year, month, day; printf("输入年 月 日: "); scanf("%d %d %d", &year, &month, &day); printf("这是该年的第 %d 天\n", days(year, month, day)); return 0;}int days(int y, int m, int d){ int sum=d; //加上前m-1月的天数 int i; for(i=1; i
解法3：后面要学习数组。然后就可以这样来了，30行之内解决问题（对数组充满期待吧）：
#include int days(int y, int m, int d);int main(){ int year, month, day; printf("输入年 月 日: "); scanf("%d %d %d", &year, &month, &day); printf("这是该年的第 %d 天\n", days(year, month, day)); return 0;}int days(int y, int m, int d){ int sum=d; int a[13]= {0,31,28,31,30,31,30,31,31,30,31,30,31}; int i; for(i=1; i 2&&((y%4==0&&y%100!=0)||y%400==0)) //若闰年，且晚于2月，加一天 sum++; return sum;}
解法4：有同学写成下面的代码，结果对，但这样的程序的确不好：
#include int days(int y, int m, int d);int main(){ int year, month, day; printf("输入年 月 日: "); scanf("%d %d %d", &year, &month, &day); printf("这是该年的第 %d 天\n", days(year, month, day)); return 0;}int days(int year, int month, int day){ int t; if(year%4==0&&year%100!=0||year%400==0) { switch(month) { case 1: t=day; break; case 2: t=day+31; break; case 3: t=day+31+29; break; case 4: t=day+31+29+31; break; case 5: t=day+31+29+31+30; break; case 6: t=day+31+29+31+30+31; break; case 7: t=day+31+29+31+30+31+30; break; case 8: t=day+31+29+31+30+31+30+31; break; case 9: t=day+31+29+31+30+31+30+31+31; break; case 10: t=day+31+29+31+30+31+30+31+31+30; break; case 11: t=day+31+29+31+30+31+30+31+31+30+31; break; case 12: t=day+31+29+31+30+31+30+31+31+30+31+30; break; } return t; } else { switch(month) { case 1: t=day; break; case 2: t=day+31; break; case 3: t=day+31+28; break; case 4: t=day+31+28+31; break; case 5: t=day+31+28+31+30; break; case 6: t=day+31+28+31+30+31; break; case 7: t=day+31+28+31+30+31+30; break; case 8: t=day+31+28+31+30+31+30+31; break; case 9: t=day+31+28+31+30+31+30+31+31; break; case 10: t=day+31+28+31+30+31+30+31+31+30; break; case 11: t=day+31+28+31+30+31+30+31+31+30+31; break; case 12: t=day+31+28+31+30+31+30+31+31+30+31+30; break; } return t; }}