[编程入门]结构体之时间设计(水题，超级笨蛋的方法）

解题思路:

注意事项:

参考代码:

#include <stdio.h>

struct time{

int year,month,day;

};

int main(){

struct time t;

int days=0;

scanf("%d%d%d",&t.year,&t.month,&t.day);

if(t.year%400==0||t.year%4==0&&t.year%100!=0){

switch(t.month){

case 1:days=t.day;break;

case 2:days=t.day+31;break;

case 3:days=t.day+31+29;break;

case 4:days=t.day+31+29+31;break;

case 5:days=t.day+31+29+31+30;break;

case 6:days=t.day+31+29+31+30+31;break;

case 7:days=t.day+31+29+31+30+31+30;break;

case 8:days=t.day+31+29+31+30+31+30+31;break;

case 9:days=t.day+31+29+31+30+31+30+31+31;break;

case 10:days=t.day+31+29+31+30+31+30+31+31+30;break;

case 11:days=t.day+31+29+31+30+31+30+31+31+30+31;break;

case 12:days=t.day+31+29+31+30+31+30+31+31+30+31+30;break;

}

}

else{

switch(t.month){

case 1:days=t.day;break;

case 2:days=t.day+31;break;

case 3:days=t.day+31+28;break;

case 4:days=t.day+31+28+31;break;

case 5:days=t.day+31+28+31+30;break;

case 6:days=t.day+31+28+31+30+31;break;

case 7:days=t.day+31+28+31+30+31+30;break;

case 8:days=t.day+31+28+31+30+31+30+31;break;

case 9:days=t.day+31+28+31+30+31+30+31+31;break;

case 10:days=t.day+31+28+31+30+31+30+31+31+30;break;

case 11:days=t.day+31+28+31+30+31+30+31+31+30+31;break;

case 12:days=t.day+31+28+31+30+31+30+31+31+30+31+30;break;

}

}

printf("%d",days);

}