按如下函数原型用函数编程解决如下的日期转换问题(要求考虑闰年的问题)

 #include <stdio.h>

int mt[13] = { 0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };

void set2(int y)
{
    if (y % 4 != 0)
    {
        mt[2] = 28;
    }
    else
    {
        if (y % 100 == 0 && y % 400 != 0)
            mt[2] = 28;
        else
            mt[2] = 29;
    }
}

void showmenu()
{
    printf("1. y/n/d -> yd\n");
    printf("2. yd -> y/n/d\n");
    printf("3. exit\n");    
}

int ymd2yd(int y, int m, int d)
{
    set2(y);
    int r = 0;
    for (int i = 0; i < m; i++)
        r += mt[i];
    r += d;
    return r;
}

void yd2ymd(int yd, int * r)
{
    set2(r[0]);
    int m = 0;
    while (yd > mt[m])
    {
        yd -= mt[m++];
    }
    r[1] = m;
    r[2] = yd;
}

int main()
{   
    char c;
    showmenu();
    while (scanf("%c", &c), 1)
    {
        switch (c)
        {
        case '1':
            int y, int m, int d;
            scanf("%d", &y);
            scanf("%d", &m);
            scanf("%d", &d);
            printf("%d\n", ymd2yd(y, m, d));
            showmenu();
            break;
        case '2':
            int yd;
            int r[3];
            scanf("%d", &r[0]);
            scanf("%d", &yd);
            yd2ymd(yd, r);
            printf("%d %d %d\n", r[0], r[1], r[2]);
            showmenu();
            break;
        case '\n':
            break;
        default:
            return 0;
        }
    }
    return 0;
}

 1. y/n/d -> yd
2. yd -> y/n/d
3. exit 1
2016 3 1
61
1. y/n/d -> yd
2. yd -> y/n/d
3. exit 1
2016 7 18
200
1. y/n/d -> yd
2. yd -> y/n/d
3. exit 1
2015 12 31
365
1. y/n/d -> yd
2. yd -> y/n/d
3. exit 2
2016 61
2016 3 1
1. y/n/d -> yd
2. yd -> y/n/d
3. exit 2
2015 61
2015 3 2
1. y/n/d -> yd
2. yd -> y/n/d
3. exit 2
2015 225
2015 8 13
1. y/n/d -> yd
2. yd -> y/n/d
3. exit 3
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