int
main()
{
	int x(0);
	int y(0);
	int d = extgcd(A, B, x, y);

	if (d == 1) {
		if (0 <= x) {
			std::printf("%d * %d + ", A, x);
		}
		else {
			std::printf("%d * (%d) + ", A, x);
		}

		if (0 <= y) {
			std::printf("%d * %d = 1\n", B, y);
		}
		else {
			std::printf("%d * (%d) = 1\n", B, y);
		}
	}

	return 0;
}
Exemple #2
0
pii extgcd(int a, int b){
	if(b == 0) return make_pair(1, 0);
	else{
		int p = a / b;
		pii q = extgcd(b, a % b);
		return make_pair(q.second, q.first - q.second * p);
	}
}
Exemple #3
0
long long extgcd( long long a,long long b,long long &x,long long &y ) {
  long long ret=a;
  if ( b ) {
    ret=extgcd( b,a%b,y,x );
    y-=( a/b )*x;
  } else x=1LL,y=0LL;
  return ret;
}
 // @desc mod mでaの逆元を求める
 template <class T> T mod_inverse( T a, T m ) {
   T x, y;
   extgcd( a, m, x, y );
   x %= m;
   while ( x < 0 )
     x += m;
   return x;
 }
Exemple #5
0
// 拓展欧几里得算法
// 有如下事实
// 1. a*x+b*y=gcd(a, b) ab不同为0,ab为整数,一定存在整数解
// 2. gcd(a, b) = gcd(b, a mod b)
//
// 那么 a*x+b*y=gcd(a, b)=gcd(b, a mod b)=b*x'+(a mod b)*y'
// 即 a*x+b*y = b*x'+(a-(a/b)*b)*y'
// a*x+b*y = a*y'+b*(x'-(a/b)*y')
// => x <- y', y <- (x'-(a/b)*y')
int extgcd(int a, int b, int& x, int& y)
{
    if (b == 0) { x = 1; y = 0; return a; };
    int gcd = extgcd(b, a % b, x, y);
    int t = y;
    y = x - (a / b)*y;
    x = t;
    return gcd;
}
int extgcd(int a, int b, int& x, int& y){
  int d = a;
  if (b != 0){
    d = extgcd(b, a%b, y, x);
    y -= (a/b) * x;
  } else {
    x = 1; y=0;
  }
  return d;
}
void extgcd(int a,int b){
	if(!b){
		if(c%a) ok=false; else ok=true,x=c/a; y=0;
		g=a>0?a:-a;
		return;
	}
	extgcd(b,a%b);
	if(!ok) return;
	int t=x; x=y; y=t-a/b*y;
}
Exemple #8
0
int
main()
{
  GEN x, y, d, u, v;
  pari_init(1000000,2);
  printf("x = "); x = gp_read_stream(stdin);
  printf("y = "); y = gp_read_stream(stdin);
  d = extgcd(x, y, &u, &v);
  pari_printf("gcd = %Ps\nu = %Ps\nv = %Ps\n", d, u, v);
  pari_close();
  return 0;
}
//(x+kb,y-ka),(x,y)
void extgcd(long long a,long long b,long long *d,long long *x,long long *y)
{
	if(!b)
	{	
		*d=a;	*x=1;	*y=0;
	}
	else
	{
		extgcd(b,a%b,d,y,x);
		*y-=(*x)*(a/b);
	}
}
//https://en.wikipedia.org/wiki/Extended_Euclidean_algorithm
int extgcd(int a, int b, int &x, int &y)
{
	if (b == 0)
	{
		x = 1;
		y = 0;
		return a;
	}
	int d = extgcd(b, a % b, x, y);
	int t = x;
	x = y;
	y = t - a / b * y;
	return d;
}
Exemple #11
0
Int divide(Int a, Int b, Int m) {
	Int g = gcd(a, m);
	if (b%g != 0)
		throw no_solution();
	Int x, y;
	extgcd(a, m, x, y);
	assert(a*x+m*y == gcd(a,m));
	Int n = x*b/g;
	Int dn = m/g;
	n -= n/dn*dn;
	if (n < 0)
		n += dn;
	return n;
}
Exemple #12
0
int main() {
  long long a,b,c,k;
  while ( ~scanf( "%I64d%I64d%I64d%I64d",&a,&b,&c,&k ) ) {
    if ( !a&&!b&&!c&&!k )break;
    long long x,y;
    long long z=extgcd( c,1LL<<k,x,y );
    long long tp=( b-a )/z;
    if ( tp*z==( b-a ) ) {
      long long t=x;
      t*=tp;
      long long mod = ( 1LL<<k )/z;
      t%=mod;
      t+=mod;
      t%=mod;
      printf( "%I64d\n",t );
    } else printf( "FOREVER\n" );
  }
  return 0;
}
Exemple #13
0
int main (void)
{
	int x,y,m,n;
	int x1,y1;
	int a,b,c;
	while(scanf("%d %d %d %d %d",&x,&y,&m,&n,&b)!=EOF)
	{
		a=m-n;
		c=y-x;
		int r=extgcd(a,b,x1,y1);
		if(c%r)
			printf("Impossible\n");
		else
		{
			x1=(c*x1/r)%b;
			if(x1<0)x1+=b;
			printf("%d\n",x1);
		}
	}
	return  0;
}
int extoclean(int x, int y, int z) {
	int ret = extgcd(x, y); int ans;
	if (z%ret) return -1;
	ans = ((long long)(z/ret)*k1%(y/ret));
	if (ans < 0) ans += y/ret; return ans;
}
int mod_inverse(int a, int m) {
        int x, y;
        extgcd(a, m, x, y);
        return (m + x % m) % m;
}
T extgcd(T a, T b, T& x, T& y) {
	T g = 1; x = 1; y = 0;
	if (b != 0) g = extgcd(b, a % b, y, x), y -= (a / b) * x;
	return g;
}
int main()
{
	long long p1, v1, p2, v2, l;
	long long t, disv, dist, td;
	long long temp;
	long long a, b, d;
	long long x, y;

	while(scanf("%lld %lld %lld %lld %lld", &p1, &p2, &v1, &v2, &l) != EOF)
	{
		while(p1 >= l)
			p1 -= l;
		while(p2 >= l)
			p2 -= l;
		if (v1 < v2)
		{
			temp = v1;
			v1 = v2;
			v2 = temp;

			temp = p1;
			p1 = p2;
			p2 = temp;
		}
		if (p1 > p2)
			dist = l - p1 + p2;
		else
			dist = p2 - p1;
		disv = v1 - v2;

		if (disv == 0)
		{
			if (dist != 0)
				printf("Impossible\n");
			else
				printf("0\n");
			continue;
		}

	//	printf("%lld %lld %lld %lld\n", p1, p2, l, dist);
		// disv * x +  (-y) * l = dist
		if (dist % gcd(disv, l) != 0)
		{
			printf("Impossible\n");
			continue;
		}

		extgcd(disv, l, &d, &a, &b);
//		printf("%lld\n", disv * a - l * b);
	//	printf("%lld %lld %lld %lld %lld %lld\n",disv, l, a, b, disv * a + l * b, d);
		x = a; y = b;
		x *= dist / d;
		y *= dist / d;
		x -= (x / (l / d)) * (l / d);
		while (x <= 0)
			x += l / d;

		printf("%lld\n", x);
	}
	return 0;
}
T invMod(T a, T mod) {
	T x, y;
	if (extgcd(a, mod, x, y) == -1) return (x + mod) % mod;
	return 0;
}
Exemple #19
0
int mod_inv(int a, int m) {
  int x, y;
  extgcd(a, m, x, y);
  return (m + x) % m;
}
Exemple #20
0
Int invMod(Int a, Int m) {
	Int x, y;
	if (extgcd(a, m, x, y) == 1) return (x + m) % m;
	else                         return 0; // unsolvable
}
Exemple #21
0
// a x + b y = gcd(a, b)
Int extgcd(Int a, Int b, Int &x, Int &y) {
	Int g = a; x = 1; y = 0;
	if (b != 0) g = extgcd(b, a % b, y, x), y -= (a / b) * x;
	return g;
}
int extgcd(int x, int y) {
	int ret, tmp;
	if (!y) { k1 = 1; k2 = 0; return x; }
	ret = extgcd(y, x%y);
	tmp = k1-x/y*k2; k1 = k2; k2 = tmp; return ret;
}