解题思路:思绪有点乱,大概就这样了
注意事项:
参考代码:
#include
int main()
{
int year,month,day;
scanf("%d %d %d",&year,&month,&day);
if(year%100==0)//此处循环是为了判断能否被100整除
{
if(year%4==0)//如果既能被100整除又能被4整除就是闰年
{
switch(month)//此处用swich和case方便列举
{
case 1:printf("%d",day);break;//每个case语句结束必须跟上break
case 2:printf("%d",31+day);break;//闰年二月29天
case 3:printf("%d",60+day);break;
case 4:printf("%d",91+day);break;
case 5:printf("d",121+day);break;
case 6:printf("%d",152+day);break;
case 7:printf("%d",182+day);break;
case 8:printf("%d",213+day);break;
case 9:printf("%d",244+day);break;
case 10:printf("%d",274+day);break;
case 11:printf("%d",305+day);break;
case 12:printf("%d",335+day);break;
}
}
if(year%4!=0)//如果能被100整除而不能被4整除就不是闰年,就得按平年计算日子数
{
switch(month)
{
case 1:printf("%d",day);break;
case 2:printf("%d",31+day);break;//平年二月28天
case 3:printf("%d",59+day);break;
case 4:printf("%d",90+day);break;
case 5:printf("d",120+day);break;
case 6:printf("%d",151+day);break;
case 7:printf("%d",181+day);break;
case 8:printf("%d",212+day);break;
case 9:printf("%d",243+day);break;
case 10:printf("%d",273+day);break;
case 11:printf("%d",304+day);break;
case 12:printf("%d",334+day);break;
}
}
}
if(year%100!=0)//这个函数是判断能否被100整除,不能被100整除则进入该函数
{
if(year%4!=0)//判断不能被100整除是否能被4整除,如果不能被4整除就按照平年计算天数
{
switch(month)
{
case 1:printf("%d",day);break;
case 2:printf("%d",31+day);break;
case 3:printf("%d",59+day);break;
case 4:printf("%d",90+day);break;
case 5:printf("d",120+day);break;
case 6:printf("%d",151+day);break;
case 7:printf("%d",181+day);break;
case 8:printf("%d",212+day);break;
case 9:printf("%d",243+day);break;
case 10:printf("%d",273+day);break;
case 11:printf("%d",304+day);break;
case 12:printf("%d",334+day);break;
}
}
if(year%4==0)//如果能被4整除就按照闰年来计数
{
switch(month)
{
case 1:printf("%d",day);break;
case 2:printf("%d",31+day);break;
case 3:printf("%d",60+day);break;
case 4:printf("%d",91+day);break;
case 5:printf("d",121+day);break;
case 6:printf("%d",152+day);break;
case 7:printf("%d",182+day);break;
case 8:printf("%d",213+day);break;
case 9:printf("%d",244+day);break;
case 10:printf("%d",274+day);break;
case 11:printf("%d",305+day);break;
case 12:printf("%d",335+day);break;
}
}
}
return 0;//函数结束
}
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