poj1061 青蛙的約會 擴展歐幾里得

（1）at+bp=c有解的條件是c%gcd(a,b)=0；

（2）求解at+bp=gcd(a,b)得x1，y1，則原式解x=x1*(c/gcd(a,b)),y=y1%(c/gcd(a,b))；

（3）問題等化爲(n-m)t+pl=x-y；

#include <stdio.h>
long long x1,y1;
long long extended(long long a,long long b)
{
    long long gcd;
    if(b==0)
    {
        x1=1;
        y1=0;
        return a;
    }
    gcd=extended(b,a%b);
    long long t=x1;
    x1=y1;
    y1=t-(a/b)*y1;
    return gcd;
}
int main()
{
    long long x,y,m,n,l;
    while(scanf("%I64d%I64d%I64d%I64d%I64d",&x,&y,&m,&n,&l)!=-1)
    {
        if(n==m)
        {
            printf("Impossible\n");
            continue;
        }
        x=(x-y)%l;
        m=(n-m)%l;
        long long gcd=extended(m,l);
        if(x%gcd!=0)
        {
            printf("Impossible\n");
            continue;
        }
        x1=x1*(x/gcd);
        l/=gcd;
        printf("%I64d\n",(x1%l+l)%l);
    }
    return 0;
}