SGU 115 Calendar

SGU_115

    类似这种日期的问题貌似还是写几个小函数处理起来更明了一些。

#include<stdio.h>
#include<string.h>
int days[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int M = 10, D = 21, W = 7;
int equals(int m, int d, int mm, int dd)
{
    if(m < mm)
        return -1;
    else if(m > mm)
        return 1;
    if(d < dd)
        return -1;
    else if(d > dd)
        return 1;
    return 0;
}
void decrease(int &m, int &d, int &w)
{
    if(d == 1)
    {
        -- m;
        d = days[m];
    }
    else
        -- d;
    w = (w + 5) % 7 + 1;
}
void increase(int &m, int &d, int &w)
{
    if(d == days[m])
        ++m, d = 1;
    else
        ++ d;
    w = w % 7 + 1;
}
int main()
{
    int i, j, k, d, m, dd, mm, ww;
    while(scanf("%d%d", &d, &m) == 2)
    {
        if(m < 1 || m > 12 || d < 1 || d > days[m])
            printf("Impossible\n");
        else
        {
            mm = M, dd = D, ww = W;
            if(equals(mm, dd, m, d) <= 0)
                for(; equals(mm, dd, m, d); increase(mm, dd, ww));
            else
                for(; equals(mm, dd, m, d); decrease(mm, dd, ww));
            printf("%d\n", ww);
        }
    }
    return 0;
}