题意：计算$sumlimits_{i=1}^n[(p{cdot }i)mod{q}]$
类欧模板题，首先作转化$sumlimits_{i=1}^n[(p{cdot}i)mod{q}]=sumlimits_{i=1}^n[p{cdot}i-leftlfloorfrac{p{cdot}i}{q} ight floor{cdot}q]$，然后只要能快速计算$sumlimits_{i=1}^nleftlfloorfrac{p{cdot}i}{q} ight floor$就行了。
记$f(a,b,c,n)=sumlimits_{i=0}^nleftlfloorfrac{ai+b}{c} ight floor$
则有$f(a,b,c,n)=left{egin{matrix}egin{aligned}&(n+1)leftlfloorfrac{b}{c} ight floor,a=0\&f(a\%c,b\%c,c,n)+frac{n(n+1)}{2}leftlfloorfrac{a}{c} ight floor+(n+1)leftlfloorfrac{b}{c} ight floor,a>=c:or:b>=c\&nleftlfloorfrac{an+b}{c} ight floor-f(c,c-b-1,a,leftlfloorfrac{an+b}{c} ight floor-1),othersend{aligned}end{matrix} ight.$
由于递推过程类似欧几里得法求gcd，因此称作类欧~~
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 const int N=1e5+10; 5 int p,q,n; 6 ll f(ll a,ll b,ll c,ll n) { 7 if(!a)return (n+1)*(b/c); 8 if(a>=c||b>=c)return f(a%c,b%c,c,n)+n*(n+1)/2*(a/c)+(n+1)*(b/c); 9 ll m=(a*n+b)/c; 10 return n*m-f(c,c-b-1,a,m-1); 11 } 12 int main() { 13 int T; 14 for(scanf("%d",&T); T--;) { 15 scanf("%d%d%d",&p,&q,&n); 16 printf("%lld ",(ll)n*(n+1)/2*p-f(p,0,q,n)*q); 17 } 18 return 0; 19 }