It's a Mod, Mod, Mod, Mod World （类欧几里得模板题

https://vjudge.net/contest/317000#problem/F

 

 

#include <iostream>
#include <cstdio>
#include <queue>
#include <algorithm>
#include <cmath>
#include <cstring>
#define inf 2147483647
#define P 998244353
#define p(a) putchar(a)
#define For(i,a,b) for(long long i=a;i<=b;++i)

using namespace std;
long long T;
long long n,a,b,c;
struct data{
    long long f,g,h;
    data calc(long long n,long long a,long long b,long long c){
         long long ac = a / c, bc = b / c, m = (a * n + b) / c, n1 = n + 1, n21 = n * 2 + 1;
         data d;
         if (a == 0) {
            d.f = bc * n1;
            d.g = bc * n * n1 /2;
            d.h = bc * bc * n1;
            return d;
        }

        if (a >= c || b >= c){
            d.f = n * n1 /2 * ac + bc * n1;
            d.g = ac * n * n1 * n21 /6 + bc * n * n1 / 2;
            d.h = ac * ac * n * n1 * n21 /6 + bc * bc * n1 + ac * bc * n * n1;
            data e = calc(n, a % c, b % c, c);
            d.h += e.h + 2 * bc * e.f + 2 * ac * e.g;
            d.g += e.g, d.f += e.f;
            return d;
        }

        data e = calc(m - 1, c, c - b - 1, a);
        d.f = n * m - e.f, d.f = d.f;
        d.g = m * n * n1 - e.h - e.f, d.g = d.g /2;
        d.h = n * m * (m + 1) - 2 * e.g - 2 * e.f - d.f;
        return d;
    }
}ans1,ans2;

void in(long long &x){
    long long y=1;char c=getchar();x=0;
    while(c<'0'||c>'9'){if(c=='-')y=-1;c=getchar();}
    while(c<='9'&&c>='0'){ x=(x<<1)+(x<<3)+c-'0';c=getchar();}
    x*=y;
}
void o(long long x){
    if(x<0){p('-');x=-x;}
    if(x>9)o(x/10);
    p(x%10+'0');
}

signed main(){
    in(T);
    while(T--){
        in(a);in(c);in(n);
        ans1=ans1.calc(n,a,0,1);
        ans2=ans2.calc(n,a,0,c);
        o(ans1.f-c*ans2.f);p('\n');
    }
    return 0;
}