poj 1061 青蛙的约会 拓展欧几里得模板

// poj 1061 青蛙的约会 拓展欧几里得模板

// 注意进行exgcd时，保证a，b是正数，最后的答案如果是负数，要加上一个膜

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>

using namespace std;
typedef long long ll;
ll gcd(ll a,ll b){
    if (b==)
        return a;
    return gcd(b,a%b);

}
ll x,y,m,n,l;

ll exgcd(ll a,ll b,ll &x,ll &y){
    if ( b ==  ){
        x = ;
        y = ;
        return a;
    }
    ll d = exgcd(b,a%b,x,y);
    ll temp = x;
    x = y;
    y = temp - a / b * y;
    return d;
}

int main(){
    //freopen("1.txt","r",stdin);
    while(scanf("%lld%lld%lld%lld%lld",&x,&y,&m,&n,&l)!=EOF){
        ll u = m - n;
        ll v = y - x;
        if (u < ){
            u = -u;
            v = -v;
        }

        ll x1,y1;
        ll d = exgcd(u,l,x1,y1);

        if (v % d){
            puts("Impossible");
            continue;
        }

        ll mod = l / d;

        x1 = x1 * ( v / d);
        printf("%lld\n",(x1%mod + mod)%mod);
    }
    return ;
}