欧几里得及拓展欧几里得
欧几里得
int gcd(int a,int b){
return (b==0)?a:gcd(b,a%b); //一条语句搞定(三元运算符)装逼,跟上面略有不同,上面做到t=0,这里做到b=0
}
拓展欧几里得
int gcd(int a,int b){
return (b==0)?a:gcd(b,a%b); //一条语句搞定(三元运算符)装逼,跟上面略有不同,上面做到t=0,这里做到b=0
}
ll lcm(ll a, ll b) {
return a / gcd(a,b) * b;
}
int exgcd(int a,int b,int &x,int &y)
{
if(b==0)
{
x=1;
y=0;
return a;
}
int r=exgcd(b,a%b,x,y);
int t=x;
x=y;
y=t-a/b*y;
return r;
}
拓展欧几里得求整数解个数
ll gcd(ll a, ll b) {
return b ? gcd(b, a%b) : a;
}
ll lcm(ll a, ll b) {
return a / gcd(a, b) * b;
}
ll extend_gcd(ll a, ll b, ll&x, ll&y) {
if (!b) {
x = 1;
y = 0;
return a;
}
ll xt = 0, yt = 0;
ll d = extend_gcd(b, a % b, xt, yt);
x = yt;
y = xt - yt * (a / b);
return d;
}
ll cal(ll a, ll b, ll n) { //计算ax+by == n的非负整数解组数
ll x = 0, y = 0, d;
d = extend_gcd(a, b, x, y);
if (n % d != 0) {
return 0;
}
x *= n / d, y *= n / d;
ll LCM = lcm(a, b);
ll t1 = LCM / a, t2 = LCM / b;
if (x < 1) {
ll num = (1 - x) / t1;
x += num * t1;
y -= num * t2;
if (x < 1) {
y -= t2;
x += t1;
}
}
if (y < 1) {
ll num = (1 - y) / t2;
y += num * t2;
x -= num * t1;
if (y < 1) {
y += t2;
x -= t1;
}
}
ll ans = x > 0 && y > 0;
if (ans) {
ans += min((x - 1) / t1, ((n - 1) / b - y) / t2);
ans += min((y - 1) / t2, ((n - 1) / a - x) / t1);
}
return ans;
}