返回：贺老师课程教学链接 项目要求
【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<stdio.h> 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<stdio.h> 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<m; i++) { switch(i) { case 2: sum+=((y%4==0&&y0!=0)||y@0==0)?29:28; break; case 4: case 6: case 9: case 11: sum+=30; break; default: sum+=31; break; } } return sum; } 解法2： #include<stdio.h> 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<m; i++) { if(i==1||i==3||i==5||i==7||i==8||i==10||i==12) sum+=31; else if (i==4||i==6||i==9||i==11) sum+=30; else sum+=((y%4==0&&y0!=0)||y@0==0)?29:28; } return sum; } 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<stdio.h> 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<m; i++) { sum+=a[i]; } if(m>2&&((y%4==0&&y0!=0)||y@0==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<stdio.h> 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&&year0!=0||year@0==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; } }