类欧几里得算法

求(sum_{x=0}^{n-1}lfloor frac{ax+b}{c} floor)

设(f(a,b,c,n)=sumlimits_{x=0}^{n-1}lfloor frac{ax+b}{c} floor)

进行化简，得

[egin{aligned} &f(a,b,c,n) \ =&sum_{x=0}^{n-1}lfloor frac{ax+b}{c} floor \ =&sum_{x=0}^{n-1}lfloor frac{(a mod c)x + (a - a mod c)x+(b mod c)+(b - b mod c)}{c} floor \ =&sum_{x=0}^{n-1}lfloor frac{(a mod c)x + (b mod c)}{c} floor + frac{(a - a mod c)x}{c} + frac{b - b mod c}{c}\ =&sum_{x=0}^{n-1}lfloor frac{(a mod c)x + (b mod c)}{c} floor + lfloor frac{a}{c} floor x + lfloor frac{b}{c} floor\ =&frac{n(n-1)}{2}lfloor frac{a}{c} floor + n lfloor frac{b}{c} floor + sum_{x=0}^{n-1}lfloor frac{(a mod c)x + (b mod c)}{c} floor \ =&frac{n(n-1)}{2}lfloor frac{a}{c} floor + n lfloor frac{b}{c} floor + f(a mod c,b mod c, c, n) end{aligned} ]

(code:)

ll f(ll a,ll b,ll c,ll n)
{
    if(n<=0) return 0;
    return n*(n-1)/2*(a/c)+n*(b/c)+f(c,(a*n+b)%c,a%c,(a%c*n+b%c)/c);
}