Kattis - itsamodmodmodmodworld It's a Mod, Mod, Mod, Mod World (类欧几里得)

题意：计算$\sum\limits_{i=1}^n[(p{\cdot }i)\bmod{q}]$

类欧模板题，首先作转化$\sum\limits_{i=1}^n[(p{\cdot}i)\bmod{q}]=\sum\limits_{i=1}^n[p{\cdot}i-\left\lfloor\frac{p{\cdot}i}{q}\right\rfloor{\cdot}q]$，然后只要能快速计算$\sum\limits_{i=1}^n\left\lfloor\frac{p{\cdot}i}{q}\right\rfloor$就行了。

记$f(a,b,c,n)=\sum\limits_{i=0}^n\left\lfloor\frac{ai+b}{c}\right\rfloor$

则有$f(a,b,c,n)=\left\{\begin{matrix}\begin{aligned}&(n+1)\left\lfloor\frac{b}{c}\right\rfloor,a=0\\&f(a\%c,b\%c,c,n)+\frac{n(n+1)}{2}\left\lfloor\frac{a}{c}\right\rfloor+(n+1)\left\lfloor\frac{b}{c}\right\rfloor,a>=c\:or\:b>=c\\&n\left\lfloor\frac{an+b}{c}\right\rfloor-f(c,c-b-1,a,\left\lfloor\frac{an+b}{c}\right\rfloor-1),others\end{aligned}\end{matrix}\right.$

由于递推过程类似欧几里得法求gcd，因此称作类欧~~

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 const int N=1e5+;
 int p,q,n;
 ll f(ll a,ll b,ll c,ll n) {
     if(!a)return (n+)*(b/c);
     if(a>=c||b>=c)return f(a%c,b%c,c,n)+n*(n+)/*(a/c)+(n+)*(b/c);
     ll m=(a*n+b)/c;
     return n*m-f(c,c-b-,a,m-);
 }
 int main() {
     int T;
     for(scanf("%d",&T); T--;) {
         scanf("%d%d%d",&p,&q,&n);
         printf("%lld\n",(ll)n*(n+)/*p-f(p,,q,n)*q);
     }
     return ;
 }