[Gym102135][G - Digital characteristic]2017-2018 8th BSUIR Open Programming Contest

description

给定a,b,m,k，产生长度为k的数组，a1=a%m,ai=(ai-1+b)%m(i>1)，将这k个数首尾相接形成一个新数，问这个数的f函数的值

‘’‘solution’’’

首先f(x)可以等价于询问x%9的值，由于10的幂模9都为1，x%9=$(\sum (a+bi)\%m)$

$$=(\sum a+bi-\lfloor \frac{a+bi}{m}\rfloor \ast m)$$
$$=\sum a+bi\%9-(m\ast \sum \lfloor \frac{a+bi}{m}\rfloor )\%9$$
$$=\sum a+bi\%9-(m\ast f(b,a,m,k-1))\%9$$

$$类欧几里得推导：f(a,b,c,n)=\sum _0^n\lfloor \frac{ai+b}{c}\rfloor$$
$当a=0时：$
$$f(a,b,c,n)=\lfloor b/c\rfloor \ast (n+1)$$
$当a>=c或b>=c时：$
$$f(a,b,c,n)=\sum _0^n\lfloor a/c\rfloor \ast i+\lfloor b/c\rfloor +\lfloor \frac{a\%c\ast i+b\%c}{c}\rfloor$$
$$=\lfloor a/c\rfloor n\ast (n+1)/2+\lfloor b/c\rfloor \ast (n+1)+f(a\%c,b\%c,c,n)$$
$当a<c且b<c时$
$$f(a,b,c,n)=\sum _{i=0}^n\sum _{j=0}^{\lfloor \frac{ai+b}{c}\rfloor -1}1$$
$$=\sum _{j=0}^{\lfloor \frac{an+b}{c}\rfloor -1}\sum _{i=0}^{n}1(j<\lfloor \frac{ai+b}{c}\rfloor )$$
$$=\sum _{j=0}^{\lfloor \frac{an+b}{c}\rfloor -1}\sum _{i=0}^{n}1(j<\lceil \frac{ai+b-c+1}{c}\rceil )$$
$$=\sum _{j=0}^{\lfloor \frac{an+b}{c}\rfloor -1}\sum _{i=0}^{n}1(i>\lfloor \frac{cj-b+c-1}{a}\rfloor )$$
$$=\sum _{j=0}^{\lfloor \frac{an+b}{c}\rfloor -1}n-\lfloor \frac{cj-b+c-1}{a}\rfloor$$
$$=n\ast \lfloor \frac{an+b}{c}\rfloor -f(c,-b+c-1,\lfloor \frac{an+b}{c}\rfloor -1)$$
同理可以求出$g(a,b,c,n)=\sum i\ast \lfloor \frac{ai+b}{c}\rfloor$
$h(a,b,c,n)=\sum \lfloor \frac{ai+b}{c}{\rfloor }^2$（平方这里需要有一个转换)$n^2=2\ast {\sum }_{i=0}^{n}i-n$

#include
#define LL long long
#define fo(i,a,b) for(LL i=a;i<=b;i++)
#define rp(i,a,b) for(LL i=a;i>=b;i--)
#define tr(t,x) for(LL t=first[x];t;t=nex[t])
using namespace std;
const LL mo=9;
LL n,m,i,t,j,k,l,x,y,z,a,b,ans,T;
LL sqr(LL n){
	return n*(n+1)/2%mo;
}
LL f(LL a,LL b,LL c,LL n){
	if (!a) return (b/c)*(n+1)%mo;
	if (a>=c || b>=c) return (f(a%c,b%c,c,n)+sqr(n)*(a/c)+(n+1)*(b/c))%mo; 
	LL t=(a*n+b)/c;
	return (t%mo*n%mo-f(c,c-b-1,a,t-1)+mo)%mo;
}
int main(){
	//freopen("data.in","r",stdin);
	scanf("%lld",&T);
	while (T--){
		scanf("%lld%lld%lld%lld",&a,&b,&m,&n);n--;
		a%=m;b%=m;
		if ((!n || ( n && !b)) && !a){
			printf("0\n"); continue;
		}
		ans=a*(n+1)%mo+sqr(n)*b%mo-m*f(b,a,m,n)%mo;
		ans=(ans+mo)%mo; 
		if (!ans) ans=9;
		printf("%lld\n",ans);
	}
}