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这是我修改了一下的代码。

Posted by lyg at 2008-10-20 11:38:47 and last updated at 2008-10-20 11:42:16
In Reply To:不能ac的O(1)代码。。。。 Posted by:Qinz at 2008-10-19 22:29:45

#include <iostream>
using std::cin;
using std::cout;
using std::endl;

long long factor,denominator;
void setAns(long long a, long long b)
{
    factor = a;
    denominator = b;
}

int main()
{
    long long n,m,k;
    cin >>m;
    while (m--)
    {
        cin >>n >>k;
        if (k<=n/2+1) setAns(1,n-k+1);
        else if (k<=5*n/6+1&&n%2==0)
        {
            if ((k-n/2)%2==0)
                setAns(2, n-(k-n/2-1));
            else
                setAns(1, n/2-(k-n/2-1)/2);
        }
        else if (k<=5*n/6+1&&n%2==1)
        {
            if ((k-n/2)%2==0)
                setAns(2, n-(k-n/2-2));
            else
                setAns(1, n/2-(k-n/2-2)/2);
        }
        else if (n%6==1)
        {
            if ((k-5*n/6)%4==2)
                setAns(2, 2*n/3-(k-5*n/6-2)/4*2+1);
            else if ((k-5*n/6)%4==3)
                setAns(3, n-(k-5*n/6-3)/4*3);
            else if ((k-5*n/6)%4==0)
                setAns(1, n/3-(k-5*n/6)/4+1);
            else
                setAns(3, n-(k-5*n/6-1)/4*3+1);
        }
        else if (n%6==2)
        {
            if ((k-5*n/6)%4==3)
                setAns(2, 2*n/3-(k-5*n/6-3)/4*2);
            else if ((k-5*n/6)%4==0)
                setAns(3, n-(k-5*n/6)/4*3+2);
            else if ((k-5*n/6)%4==1)
                setAns(1, n/3-(k-5*n/6-1)/4+1);
            else
                setAns(3, n-(k-5*n/6-2)/4*3);
        }
        else if (n%6==3)
        {
            if ((k-5*n/6)%4==0)
                setAns(2, 2*n/3-(k-5*n/6)/4*2+1);
            else if ((k-5*n/6)%4==1)
                setAns(3, n-(k-5*n/6-1)/4*3+1);
            else if ((k-5*n/6)%4==2)
                setAns(1, n/3-(k-5*n/6-2)/4);
            else
                setAns(3, n-(k-5*n/6-3)/4*3-1);
        }
        else if (n%6==4)
        {
            if ((k-5*n/6)%4==1)
                setAns(2, 2*n/3-(k-5*n/6-1)/4*2+1);
            else if ((k-5*n/6)%4==2)
                setAns(3, n-(k-5*n/6-2)/4*3);
            else if ((k-5*n/6)%4==3)
                setAns(1, n/3-(k-5*n/6-3)/4);
            else
                setAns(3, n-(k-5*n/6-4)/4*3-2);
        }
        else if (n%6==5)
        {
            if ((k-5*n/6)%4==3)
                setAns(2, 2*n/3-(k-5*n/6-3)/4*2);
            else if ((k-5*n/6)%4==0)
                setAns(3, n-(k-5*n/6)/4*3+2);
            else if ((k-5*n/6)%4==1)
                setAns(1, n/3-(k-5*n/6-1)/4+1);
            else
                setAns(3, n-(k-5*n/6-2)/4*3);
        }
        else
        {
            if ((k-5*n/6)%4==3)
                setAns(2, 2*n/3-(k-5*n/6-3)/4*2-1);
            else if ((k-5*n/6)%4==0)
                setAns(3, n-(k-5*n/6)/4*3+1);
            else if ((k-5*n/6)%4==1)
                setAns(1, n/3-(k-5*n/6-1)/4);
            else
                setAns(3, n-(k-5*n/6-2)/4*3-1);
        }
        cout <<factor <<"/" <<denominator <<endl;
    }
    return 0;
}

Followed by:

成都网赛在这题耗了太多时间，以至最后看到1004有思路但没时间写。。。我感觉我还有很长的acm之路得走，感谢各位大牛的指导 (0) Qinz 2008-10-20 21:15:33

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Content:

> #include <iostream>
> using std::cin;
> using std::cout;
> using std::endl;
> 
> long long factor,denominator;
> void setAns(long long a, long long b)
> {
>     factor = a;
>     denominator = b;
> }
> 
> int main()
> {
>     long long n,m,k;
>     cin >>m;
>     while (m--)
>     {
>         cin >>n >>k;
>         if (k<=n/2+1) setAns(1,n-k+1);
>         else if (k<=5*n/6+1&&n%2==0)
>         {
>             if ((k-n/2)%2==0)
>                 setAns(2, n-(k-n/2-1));
>             else
>                 setAns(1, n/2-(k-n/2-1)/2);
>         }
>         else if (k<=5*n/6+1&&n%2==1)
>         {
>             if ((k-n/2)%2==0)
>                 setAns(2, n-(k-n/2-2));
>             else
>                 setAns(1, n/2-(k-n/2-2)/2);
>         }
>         else if (n%6==1)
>         {
>             if ((k-5*n/6)%4==2)
>                 setAns(2, 2*n/3-(k-5*n/6-2)/4*2+1);
>             else if ((k-5*n/6)%4==3)
>                 setAns(3, n-(k-5*n/6-3)/4*3);
>             else if ((k-5*n/6)%4==0)
>                 setAns(1, n/3-(k-5*n/6)/4+1);
>             else
>                 setAns(3, n-(k-5*n/6-1)/4*3+1);
>         }
>         else if (n%6==2)
>         {
>             if ((k-5*n/6)%4==3)
>                 setAns(2, 2*n/3-(k-5*n/6-3)/4*2);
>             else if ((k-5*n/6)%4==0)
>                 setAns(3, n-(k-5*n/6)/4*3+2);
>             else if ((k-5*n/6)%4==1)
>                 setAns(1, n/3-(k-5*n/6-1)/4+1);
>             else
>                 setAns(3, n-(k-5*n/6-2)/4*3);
>         }
>         else if (n%6==3)
>         {
>             if ((k-5*n/6)%4==0)
>                 setAns(2, 2*n/3-(k-5*n/6)/4*2+1);
>             else if ((k-5*n/6)%4==1)
>                 setAns(3, n-(k-5*n/6-1)/4*3+1);
>             else if ((k-5*n/6)%4==2)
>                 setAns(1, n/3-(k-5*n/6-2)/4);
>             else
>                 setAns(3, n-(k-5*n/6-3)/4*3-1);
>         }
>         else if (n%6==4)
>         {
>             if ((k-5*n/6)%4==1)
>                 setAns(2, 2*n/3-(k-5*n/6-1)/4*2+1);
>             else if ((k-5*n/6)%4==2)
>                 setAns(3, n-(k-5*n/6-2)/4*3);
>             else if ((k-5*n/6)%4==3)
>                 setAns(1, n/3-(k-5*n/6-3)/4);
>             else
>                 setAns(3, n-(k-5*n/6-4)/4*3-2);
>         }
>         else if (n%6==5)
>         {
>             if ((k-5*n/6)%4==3)
>                 setAns(2, 2*n/3-(k-5*n/6-3)/4*2);
>             else if ((k-5*n/6)%4==0)
>                 setAns(3, n-(k-5*n/6)/4*3+2);
>             else if ((k-5*n/6)%4==1)
>                 setAns(1, n/3-(k-5*n/6-1)/4+1);
>             else
>                 setAns(3, n-(k-5*n/6-2)/4*3);
>         }
>         else
>         {
>             if ((k-5*n/6)%4==3)
>                 setAns(2, 2*n/3-(k-5*n/6-3)/4*2-1);
>             else if ((k-5*n/6)%4==0)
>                 setAns(3, n-(k-5*n/6)/4*3+1);
>             else if ((k-5*n/6)%4==1)
>                 setAns(1, n/3-(k-5*n/6-1)/4);
>             else
>                 setAns(3, n-(k-5*n/6-2)/4*3-1);
>         }
>         cout <<factor <<"/" <<denominator <<endl;
>     }
>     return 0;
> }
>

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