类欧几里得小结
前言：网上看了一下这个知识点，最基本的那个式子会了之后，自己推了剩下的两个式子，发现其实还是很好推的
前置公式：
$x < \left \lfloor \frac{n}{y} \right \rfloor \Leftrightarrow x < \left \lceil \frac{n-y+1}{y} \right \rceil$
$x < \left \lceil \frac{n}{y} \right \rceil \Leftrightarrow x * y < n$
$x * y > n \Leftrightarrow x > \left \lfloor \frac{n}{y} \right \rfloor$
类欧几里得：
给定 $n$, $a$, $b$, $c$，求下列三个算式的值
$f(n, a, b, c) = \sum_{i=0}^n\left \lfloor \frac{a*i+b}{c} \right \rfloor$
$g(n, a, b, c) = \sum_{i=0}^n i * \left \lfloor \frac{a*i+b}{c} \right \rfloor$
$h(n, a, b, c) = \sum_{i=0}^n (\left \lfloor \frac{a*i+b}{c} \right \rfloor)^2$
本文中作如下约定：
$m=\left \lfloor \frac{a*n+b}{c} \right \rfloor$
求$f(n, a, b, c)$
$Case\ 1$:当$a \geqslant c \ ||\ b \geqslant c$时 $$ \begin{align*}f(n, a, b, c) &= \sum_{i=0}^n\left \lfloor \frac{a*i+b}{c} \right \rfloor \\ &=\sum_{i=0}^n(\left \lfloor \frac{(a\%c)*i+(b\%c)}{c} \right \rfloor + \left \lfloor \frac{a}{c} \right \rfloor*i + \left \lfloor \frac{b}{c} \right \rfloor ) \\ &= \sum_{i=0}^n\left \lfloor \frac{(a\%c)*i+(b\%c)}{c} \right \rfloor + \left \lfloor \frac{a}{c} \right \rfloor \sum_{i=0}^ni + \left \lfloor \frac{b}{c} \right \rfloor * (n + 1) \\ &=f(n, a\%c, b\%c, c) + \left \lfloor \frac{a}{c} \right \rfloor * \frac{n*(n+1)}{2} + \left \lfloor \frac{b}{c} \right \rfloor * (n + 1) \end{align*}$$
$Case\ 2$:当$a < c \ \&\&\ b < c \ \&\&\ a \neq 0$时 $$\begin{align*}f(n, a, b, c) &= \sum_{i=0}^n\left \lfloor \frac{a*i+b}{c} \right \rfloor \\ &=\sum_{i=0}^n \sum_{j=0}^{\left \lfloor \frac{a*i+b}{c} \right \rfloor - 1}1 \\ &=\sum_{j=0}^{m-1} \sum_{i=0}^n[j<\left \lfloor \frac{a*i+b}{c} \right \rfloor] \\ &= \sum_{j=0}^{m-1}\sum_{i=0}^n[j<\left \lceil \frac{a*i+b-c+1}{c} \right \rceil] \\ &= \sum_{j=0}^{m-1}\sum_{i=0}^n[c*j < a*i+b-c+1] \\&= \sum_{j=0}^{m-1}\sum_{i=0}^n[a*i > c*j-b+c-1] \\ &=\sum_{j=0}^{m-1} \sum_{i=0}^n[i>\left \lfloor \frac{c*j-b+c-1}{a} \right \rfloor] \\ &= \sum_{j=0}^{m-1}(n-\left \lfloor \frac{c*j-b+c-1}{a} \right \rfloor)\\ &= n*m-f(m-1,c,-b+c-1,a)\end{align*}$$
$Case\ 3$:当$a < c \ \&\&\ b < c \ \&\&\ a = 0$时 $$f(n, a, b, c) = 0$$
记$A(j)=\left \lfloor \frac{c*j-b+c-1}{a} \right \rfloor$
下面推导过程中与上面相似的步骤会省略
求$g(n, a, b, c)$
$Case\ 1$:当$a \geqslant c \ ||\ b \geqslant c$时 $$ \begin{align*}g(n, a, b, c)&= \sum_{i=0}^n i*\left \lfloor \frac{a*i+b}{c} \right \rfloor \\ &=\sum_{i=0}^n(i*\left \lfloor \frac{(a\%c)*i+(b\%c)}{c} \right \rfloor + \left \lfloor \frac{a}{c} \right \rfloor*i^2 + \left \lfloor \frac{b}{c} \right \rfloor * i ) \\ &= \sum_{i=0}^n i * \left \lfloor \frac{(a\%c)*i+(b\%c)}{c} \right \rfloor + \left \lfloor \frac{a}{c} \right \rfloor \sum_{i=0}^n i^2 + \left \lfloor \frac{b}{c} \right \rfloor \sum_{i=0}^n i \\ &=g(n, a\%c, b\%c, c) + \left \lfloor \frac{a}{c} \right \rfloor * \frac{n*(n+1)*(2n+1)}{6} + \left \lfloor \frac{b}{c} \right \rfloor * \frac{n*(n+1)}{2}\end{align*} $$
$Case\ 2$:当$a < c \ \&\&\ b < c \ \&\&\ a \neq 0$时 $$ \begin{align*}g(n, a, b, c) &= \sum_{i=0}^ni*\left \lfloor \frac{a*i+b}{c} \right \rfloor \\ &=\sum_{i=0}^n \sum_{j=0}^{\left \lfloor \frac{a*i+b}{c} \right \rfloor - 1}i \\ &=\sum_{j=0}^{m-1} \sum_{i=0}^ni*[j<\left \lfloor \frac{a*i+b}{c} \right \rfloor] \\ &=\sum_{j=0}^{m-1} \sum_{i=0}^ni*[i>\left \lfloor \frac{c*j-b+c-1}{a} \right \rfloor] \\ &= \sum_{j=0}^{m-1}(\frac{n*(n+1)}{2} -\frac{A(j)*(A(j)+1)}{2})\\ &= \frac{m*n*(n+1)}{2}- \frac{1}{2}\sum_{j=0}^{m-1}(A(j))^2- \frac{1}{2}\sum_{j=0}^{m-1}A(j)\\ &=\frac{m*n*(n+1)}{2}-\frac{1}{2}h(m-1, c, -b+c-1,a)-\frac{1}{2}f(m-1, c,-b+c-1, a)\end{align*}$$
$Case\ 3$:当$a < c \ \&\&\ b < c \ \&\&\ a = 0$时 $$g(n, a, b, c) = 0$$
求$h(n, a, b, c)$
$Case\ 1$:当$a \geqslant c \ ||\ b \geqslant c$时 $$\begin{align*}h(n, a, b, c) &= \sum_{i=0}^n(\left \lfloor \frac{a*i+b}{c} \right \rfloor)^2 \\ &=\sum_{i=0}^n(\left \lfloor \frac{(a\%c)*i+(b\%c)}{c} \right \rfloor + \left \lfloor \frac{a}{c} \right \rfloor*i + \left \lfloor \frac{b}{c} \right \rfloor )^2 \\ &= \sum_{i=0}^n(\left \lfloor \frac{(a\%c)*i+(b\%c)}{c} \right \rfloor)^2 +2\left \lfloor \frac{a}{c} \right \rfloor\sum_{i=0}^ni \left \lfloor \frac{(a\%c)*i+(b\%c)}{c} \right \rfloor +2\left \lfloor \frac{b}{c} \right \rfloor\sum_{i=0}^n \left \lfloor \frac{(a\%c)*i+(b\%c)}{c} \right \rfloor+ (\left \lfloor \frac{a}{c} \right \rfloor)^2 \sum_{i=0}^ni^2 + (\left \lfloor \frac{b}{c} \right \rfloor)^2 *(n+1)+2\left \lfloor \frac{a}{c} \right \rfloor\left \lfloor \frac{b}{c} \right \rfloor \sum_{i=0}^ni \\ &= h(n, a\%c, b\%c, c)+2\left \lfloor \frac{a}{c} \right \rfloor g(n,a\%c, b\%c, c)+2\left \lfloor \frac{b}{c} \right \rfloor f(n, a\%c, b\%c, c)+(\left \lfloor \frac{a}{c} \right \rfloor)^2*\frac{n*(n+1)*(2n+1)}{6}+ (\left \lfloor \frac{b}{c} \right \rfloor)^2 *(n+1)+2\left \lfloor \frac{a}{c} \right \rfloor\left \lfloor \frac{b}{c} \right \rfloor*\frac{n*(n+1)}{2}\end{align*}$$
$Case\ 2$:当$a<c\ \&\&\ b<c\ \&\&\ a\neq 0$时 $$\begin{align*}h(n,a,b,c) &=\sum_{i=0}^n(\left \lfloor \frac{a*i+b}{c} \right \rfloor)^2\\ &=\sum_{i=0}^n \sum_{j=0}^{\left \lfloor \frac{a*i+b}{c} \right \rfloor -1}\sum_{k=0}^{\left \lfloor \frac{a*i+b}{c} \right \rfloor -1}1\\ &=\sum_{j=0}^{m-1}\sum_{k=0}^{m-1}\sum_{i=0}^n[j<\left \lfloor \frac{a*i+b}{c} \right \rfloor\&\&k<\left \lfloor \frac{a*i+b}{c} \right \rfloor]\\ &=\sum_{j=0}^{m-1}\sum_{k=0}^{m-1}\sum_{i=0}^n[i>A(j)\&\&i>A(k))]\\ &= \sum_{j=0}^{m-1}\sum_{k=0}^{m-1}\sum_{i=0}^n[i>A(max(j,k))]\\ &=2\sum_{j=0}^{m-1}\sum_{k=0}^{j-1}\sum_{i=0}^n[i>A(j)] +\sum_{j=0}^{m-1}\sum_{i=0}^n[i>A(j)]\\ &= 2\sum_{j=0}^{m-1}j*(n-A(j))+\sum_{j=0}^{m-1} (n-A(j))\\ &=2n\sum_{j=0}^{m-1}j-2\sum_{j=0}^{m-1}j*A(j)+m*n-\sum_{j=0}^{m-1}A(j)\\ &=2n*\frac{(m-1)*m}{2}+m*n-2g(m-1, c,-b+c-1,a)-f(m-1,c,-b+c-1,a)\\ &=n*m^2-2g(m-1, c,-b+c-1,a)-f(m-1,c,-b+c-1,a)\end{align*}$$
$Case\ 3$:当$a < c \ \&\&\ b < c \ \&\&\ a = 0$时 $$h(n, a, b, c) = 0$$
代码：(洛谷P5170)
#include<cstdio> #include<iostream> #include<algorithm> #include<cstring> using namespace std; typedef long long ll; const int mod = 998244353; inline int read() { int ret, f=1; char c; while((c=getchar())&&(c<'0'||c>'9'))if(c=='-')f=-1; ret=c-'0'; while((c=getchar())&&(c>='0'&&c<='9'))ret=(ret<<3)+(ret<<1)+c-'0'; return ret*f; } int T, n, inv2, inv6; struct node{ ll fir, sec, thr; node(){} node(ll x, ll y, ll z) { fir = x; sec = y; thr = z; } }ans; ll add(ll x, ll y) { return x + y < mod? x + y: x + y - mod; } ll rdc(ll x, ll y) { return x - y < 0? x - y + mod: x - y; } ll qpow(ll x, int y) { ll ret = 1; while(y) { if(y&1) ret = ret * x % mod; x = x * x % mod; y >>= 1; } return ret; } ll calc1(ll x) { return (x * (x + 1) % mod) * inv2 % mod; } ll calc2(ll x) { return ((x * (x + 1) % mod) * (2 * x + 1) % mod) * inv6 % mod; } node dfs(ll x, ll a, ll b, ll c) { if(x < 0) return node(0, 0, 0); if(a >= c || b >= c) { node ret, res = dfs(x, a % c, b % c, c); ret.fir = (a / c) * calc1(x) % mod; ret.fir = add(ret.fir, (x + 1) * (b / c) % mod); ret.fir = add(ret.fir, res.fir); ret.sec = add((a / c) * calc2(x) % mod, (b / c) * calc1(x) % mod); ret.sec = add(ret.sec, res.sec); ret.thr = res.thr; ret.thr = add(ret.thr, (2 * (a / c) % mod) * res.sec % mod); ret.thr = add(ret.thr, (2 * (b / c) % mod) * res.fir % mod); ret.thr = add(ret.thr, (2 * (a / c) * (b / c) % mod) * calc1(x) % mod); ret.thr = add(ret.thr, ((a / c) * (a / c) % mod) * calc2(x) % mod); ret.thr = add(ret.thr, ((b / c) * (b / c) % mod) * (x + 1) % mod); return ret; } if(a == 0 || x == 0) return node(0, 0, 0); ll m = (a * x + b) / c; node res = dfs(m - 1, c, - b + c - 1, a), ret; ret.fir = rdc(m * x % mod, res.fir); ret.sec = add(res.thr, res.fir); ret.sec = ret.sec * inv2 % mod; ret.sec = rdc(m * calc1(x) % mod, ret.sec); ret.thr = (x * m % mod) * m % mod; ret.thr = rdc(ret.thr, add(2 * res.sec % mod, res.fir)); return ret; } int main() { T = read(); int a, b, c; inv2 = (mod + 1) >> 1; inv6 = qpow(6, mod - 2); while(T --) { n = read(); a = read(); b = read(); c = read(); ans = dfs(n, a, b, c); printf("%lld %lld %lld\n", ans.fir, ans.thr, ans.sec); } return 0; }
