uva 10277 Boastin' Red Socks

原题：
You have a drawer that is full of two kinds of socks: red and black. You know that there are at least 2 socks, and not more than 50000. However, you do not know how many there actually are, nor do you know how many are red, or how many are black. (Your mother does the laundry!)
You have noticed, though, that when you reach into the drawer each morning and choose two socks to wear (in pitch darkness, so you cannot distinguish red from black), the probability that you pick two red socks is exactly p/q, where 0 < q and 0 ≤ p ≤ q. From this, can you determine how many socks of each colour are in your drawer? There may be multiple solutions — if so, pick the solution with the fewest total number of socks, but still allowing you to wear a couple of same color socks.
Input
Input consists of multiple problems, each on a separate line. Each problem consists of the integers p and q separated by a single space. Note that p and q will both fit into an unsigned long integer.
Input is terminated by a line consisting of two zeroes.
Output
For each problem, output a single line consisting of the number of red socks and the number of black
socks in your drawer, separated by one space. If there is no solution to the problem, print ‘impossible’.
Sample Input
1 2
6 8
12 2499550020
56 789
0 0
Sample Output
3 1
7 1
4 49992
impossible
大意：
给你两个数p/q表示在一堆红色和白色袜子中取出两只袜子为为红色袜子的概率，告诉你总袜子数不超过50000,现在问你红袜子和黑袜子有多少个？如果答案有多个，找出总数最小的那个。

#include <bits/stdc++.h>using namespace std;//fstream in,out;long long q,p;long long gcd(long long a,long long b){    return b==0 ? a:gcd(b,a%b);}bool IsSqure(long long a){    long long tmp=sqrt(a);    return tmp*tmp==a;}int main(){    ios::sync_with_stdio(false);    long long tot,red,tmp;    while(cin>>p>>q,q+p)    {        if(p==0)        {            cout<<"0 2"<<endl;            continue;        }        if(p==q)        {            cout<<"2 0"<<endl;            continue;        }        long long g=gcd(p,q);        q/=g;        p/=g;        bool flag=0;        for(long long tot=2;tot<=50000;++tot)        {            if((tot*(tot-1))%q==0)            {                tmp=tot*(tot-1)/q;                if(IsSqure(4*p*tmp+1))                {                    tmp=sqrt(4*p*tmp+1);                    if((1+tmp)%2==0)                    {                        red=(1+tmp)/2;                        cout<<red<<" "<<tot-red<<endl;                        flag=1;                        break;                    }                }            }        }        if(!flag)            cout<<"impossible"<<endl;    }    return 0;}

解答：
这题应该不能算是数学题吧，首先枚举总数，然后根据方程判断能否整除和开根号之类的，如果都满足就输出结果就可以了。