关于#输出格式#的问题，如何解决？

题要求编写程序计算某年某月某日是该年中的第几天。

输入格式:
输入在一行中按照格式“y/m/d”（即“年/月/日”）给出日期。注意：闰年的判别条件是该年年份能被4整除但不能被100整除、或者能被400整除。闰年的2月有29天。

输出格式:
在一行输出日期是该年中的第几天。

输入样例1:
2009/03/02
输出样例1:
61

供参考：

#include<stdio.h>
int isLeap(int y)
{
    return ((y % 4 == 0 && y % 100 != 0) || (y % 400 == 0));
}
int Getdaysofmonth(int y, int m)
{
    int days[] = { 31,28,31,30,31,30,31,31,30,31,30,31 };
    if (m == 2 && isLeap(y))  return days[m - 1] + 1;
    return days[m - 1];
}
int main()
{
    int y, m, d, sum, mm;
    //printf("请输入年月日(yyyy/mm/dd)：\n");
    scanf("%d/%d/%d", &y, &m, &d);
    for (mm = 1, sum = 0; mm < m; sum += Getdaysofmonth(y, mm), mm++);
    sum += d;
    printf("%d", sum);
    return 0;
}

#include <stdio.h>
#include<math.h>

 
int main()
{        int a[13]={0,31,28,31,30,31,30,31,31,30,31,30,31},year,month,day,sum=0,i;
        scanf("%d/%d/%d",&year,&month,&day);
        if(year%400==0||year%4==0&&year%100!=0)
            a[0]=day,a[2]=29;
        else a[0]=day;      
        for(i=0;i<month;i++)
            sum+=a[i];
        printf("%d ",sum);

}

#include <stdio.h>

int main(int argc, char* argv[]) {
    int yearday[12] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    int year, month, day;
    int num = 0;
    scanf("%d/%d/%d", &year, &month, &day);
    if(month == 1) {
        printf("%d\n", day);
        return 0;
    }
    if((year % 4 == 0 && year % 100 != 0) || (year % 400 == 0))
        yearday[1] = 29;
    while(num < month)
        num += yearday[num ++];
    num += day;
    printf("%d\n", num);
    return 0;
}