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Esspe-Peasee

Esspe-Peasee is an ancient game played by children throughout the land of Acmania. The rules are simple:

A player simply quibs the yorba at the kwonk. If the yorba hurms the kwonk the player gets a foom. If the yorba hurfs the kwonk the player gets a foob.

The objective is to get a twob with as few quibs as possible.

Every group of children has its own opinion regarding the value of a foom, the value of a foob, and the value of a twob. However, everyone agrees that a foob is worth more than a foom, and that a twob is worth more than a foob. You may assume that a foom and a foob can each be represented by a 32 bit integer, and a twob can be represented by a 64 bit integer.

Input

You will be given a number of game instances to solve. Each instance is specified by 3 non-negative integers that represent the value of a foom, a foob and a twob, respectively. The final line contains three 0's and should not be processed.

Output

For each instance your program should print `A fooms and B foobs for a twob!', on a line by itself as shown in the samples below, where the value of ``A" fooms plus ``B" foobs add up to a twob, and the sum of ``A" and ``B" is as small as possible. ``fooms" and ``foobs" should be appropriately pluralised, as shown in ``Sample Output" below.

If there is no such pair you should print out the age-old chant: `Unquibable!'

Sample Input

1 6 15
7 9 22
7 9 32
0 9 18
2 5 9
0 0 0

Sample Output

3 fooms and 2 foobs for a twob!
Unquibable!
2 fooms and 2 foobs for a twob!
0 fooms and 2 foobs for a twob!
2 fooms and 1 foob for a twob!

扩展欧几里得算法不再累赘，网上各种大神讲解。orz

顺便总结一下

ax+by=c 若有解，即c%gcd(a,b)==0，以下均为有解情况：

　　若c=1 && gcd(a,b)==1

　　　　　　特解 (x0 , y0)

　　　　　　通解 (x0+b*t , y0-a*t)

　　若c==_c*gcd(a,b)

　　　　　　原方程左右同除gcd(a,b)可简化为 _ax+_by=_c gcd(_a,_b)==1 故转化为上述情况 _ax+_by=1

　　　　　　求得特解为 (x0*_c , y0*_c)

　　　　　　故通解为 (x0*_c+_b*t, y0*_c-_a*t) 即 (x0*c/gcd(a,b)+b/gcd(a,b)*t , y0*c/gcd(a,b)-a/gcd(a,b)*t)

这题值得注意而且经常需要用到的地方，就是x,y>0且保证x+y最小

　　若需x>0 可直接

　　　　x=(x*c%b+b)%b即可找出最小的正数x

　　　　y=(c-a*x)/b即可求得对应y

　　　　　　若y<0 则不可能出现x,y同时>0的情况，因为x已经是最小的正数，若减小，则x为负，若增大，则y会减小，y为负

#include <cstdio>
#include <cstring>
long long a,b,c,d,x,y;

long long exgcd(long long a, long long b, long long &x, long long &y)
{
    if(b==0)
    {
        x=1;
        y=0;
        return a;
    }
    long long ret=exgcd(b, a%b, x, y);
    long long ty=y;
    y=x-a/b*y;
    x=ty;
    return ret;
}

int main()
{
    while(scanf("%lld%lld%lld",&a,&b,&c)!=EOF &&(a || b || c))
    {
        long long d=exgcd(a,b,x,y);

        if(c%d!=0)
            printf("Unquibable!
");
        else
        {
            a=a/d;
            b=b/d;
            c=c/d;
            x=(((x%b)*(c%b)%b)+b)%b;//x刚好大于0 即x的前一个就已经小于0 若使得y小于0 说明无解
            y=(c-a*x)/b;
            if(y<0)
            {
                printf("Unquibable!
");
                continue;
            }
            if(x==1)
            {
                printf("1 foom and ");
                if(y==1) printf("1 foob for a twob!
");
                else printf("%lld foobs for a twob!
",y);
            }else
            {
                printf("%lld fooms and ",x);
                if(y==1) printf("1 foob for a twob!
");
                else printf("%lld foobs for a twob!
",y);
            }

        }
    }
    return 0;
}