বুধবার, ৯ ডিসেম্বর, ২০১৫

LOJ 1010 Knights

Problem id :    LOJ 1010


To maximize the number of knights we have to place knights in a particular pattern.


Code :



 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
#include<stdio.h>
int main()
{

    int tc,cas = 0 ;
    int ans = 0;
    scanf("%d",&tc);
    while(tc--)
    {
        int m,n;
        scanf("%d %d",&m,&n);
        //ans = solve(m,n);
        //ROW / COL = 1

        if(m==1)
            ans = n;
        else if(n==1)
            ans = m;

        //ROW / COL = 2
        else if(m==2)
        {
            ans = 0;
            for(int i=1,cn=1; i<=n; i+=4,cn++)
            {

                //    if(cn&1)
                {
                    if(i==n)
                        ans +=2;
                    else
                        ans += 4;
                }
            }
        }
        else if(n==2)
        {
            ans = 0;
            for(int i=1,cn=1; i<=m; i+=4,cn++)
            {

                //     if(cn&1)
                {
                    if(i==m)
                        ans +=2;
                    else
                        ans += 4;
                }
            }
        }

//ROW / COL > 2
        else if(m&1)//odd
        {
            if(n&1)//odd
                ans= (m/2 + 1)*(n/2 + 1) + (m/2)*(n/2);
            else //m odd n even
                ans= (m/2 + 1)*(n/2) + (m/2)*(n/2);
        }
        else
        {
            if(n&1)//odd
                ans= (m/2)*(n/2 + 1) + (m/2)*(n/2);
            else
                ans= (m/2)*(n/2) + (m/2)*(n/2);
        }
        printf("Case %d: %d\n",++cas,ans);
    }

    return 0;
}

কোন মন্তব্য নেই:

একটি মন্তব্য পোস্ট করুন