题解 | #计算日期到天数转换#

#include <stdio.h>

int main() {
    int year,month,day,sum=0;
    scanf("%d%d%d",&year,&month,&day);
    switch(month)
    {
        case 1:sum=day;break;
        case 2:sum=31+day;break;
        case 3:sum=31+28+day;break;
        case 4:sum=31+28+31+day;break;
        case 5:sum=31+28+31+30+day;break;
        case 6:sum=31+28+31+30+31+day;break;
        case 7:sum=31+28+31+30+31+30+day;break;
        case 8:sum=31+28+31+30+31+30+31+day;break;
        case 9:sum=31+28+31+30+31+30+31+31+day;break;
        case 10:sum=31+28+31+30+31+30+31+31+30+day;break;
        case 11:sum=31+28+31+30+31+30+31+31+30+31+day;break;
        case 12:sum=31+28+31+30+31+30+31+31+30+31+30+day;break;
    }
    if(((year%4==0&&year%100!=0)||(year%400==0))&&(month>2))
    {
        sum+=1;
        printf("%d\n",sum);
    }
    else
    printf("%d\n",sum);
    return 0;
}