はじめに：オンラインは、この知識で、後だろう基本的な式を見て、彼は2つの式の残りの部分を押し、それはまだ良いプッシュを発見されました
前置公式：
$ X <\左\ lfloor \ FRAC {n}は{Y} \右\ rfloor \ Leftrightarrow X <\ \ lceilを左\ FRAC {N-Y + 1} {Y} \右\ rceil $
$ X <\ \ lceil \ FRAC {n}は{Y} \右\ rceil \ Leftrightarrow X * Y <N $を残し
$ X * Y> N \ Leftrightarrow X> \左\ lfloor \ FRAC {n}は{Y} \右\ rfloor $
ユークリッドカテゴリ：
$ N $、$ $、$、$ C $ B $、次の三つの式の評価を考えます
$ F（N、A、B、C）= \ sum_ {I = 0} ^ N \左\ lfloor \ FRAC {* I + B}、{C} \右\ rfloor $
$ G（N、A、B、C）= \ sum_ {iは= 0} ^ NI * \は\ lfloor \ FRACを左{* I + B}、{C} \右\ rfloor $
$ H（N、A、B、C）= \ sum_ ^ {iが0 = N} ^ 2 $（\ \ lfloor \ FRAC {* I + B}、{C} \ \右rfloorを左）
次のように表記は、本明細書中で使用されます：
$ M = \左\ lfloor \ FRAC {* N + B}、{C} \右\ rfloor $
要求の$ F（N、A、B、C）$
$ケース\ 1 $：当$ A \ geqslant C \ || \ B \ geqslant C $时 $$ \開始{ALIGN *} F（N、A、B、C）＆= \ sum_ {I = 0} ^ N \左\ lfloor \ FRAC {* I + B}、{C} \右\ rfloor \\＆= \ sum_ {I = 0} ^ N（\ \ lfloor \ FRAC {（\％Cの左側）* I + （B \％C）}、{C} \右\ rfloor + \ * I + \ rfloor \ \右\ lfloor \ FRAC {B}、{C}を左\ rfloor \ lfloor \ FRAC {} {C} \右を左\ lfloor \ FRAC左）\\＆= \ sum_ {i = 0} ^ N \ \ lfloorを左\ FRAC {（\％のC）* I +（B \％C）}、{C} \右\ rfloor + \ {} {C} \右\ rfloor \ sum_ {I = 0} ^ NI + \左\ lfloor \ FRAC {B}、{C} \右\ rfloor *（N + 1）\\＆= F（N、 \％のC、B \％のC、C）+ \ \ lfloor \ FRACを左{} {C} rfloor \ \右* \ FRAC {N *（N + 1）} {2} +は\ \ lfloorを左\ FRAC {B}、{C} \右\ rfloor *（N + 1）\端{ALIGN *} $$
$ケース\ 2 $：当$ <C \ \＆\＆\ B <C \ \＆\＆\ A \ NEQ 0 $时$$ \ {ALIGN *} Fを開始する（N、A、B、C） ＆= \ sum_ ^ {iが0 =} N \左\ lfloor \ FRAC {* I + B}、{C} \右\ rfloor \\＆= \ sum_ {I = 0} ^ nが\ sum_ {J = 0 } ^ {C} \右\ rfloor {\ \ lfloor \ FRAC {* I + B左} - 1} 1 \\＆= \ sum_ {J = 0} ^ {M-1} \ sum_ {I = 0を} ^ N \\＆= \ sum_ {J = 0} ^ {M-1} \ sum_ {[jが<\ \ lfloor \ FRAC {* I + B}、{C} \右\ rfloor左]をiが0 = } ^ N \\＆= \ sum_ {J = 0} ^ {M-1} \ sum_ [jが<\ FRAC {* I + B-C + 1} {C} \右\ rceil \ \ lceilを残し] ^ N [C * J <* I + B-C + 1] \\＆= \ sum_ {J = 0} ^ {M-1} \ sum_ {I = 0} ^ N [A {iが= 0} * iが> C * J-B + C-1] \\＆= \ sum_ {J = 0} ^ {M-1} \ sum_ {I = 0} ^ N [I> \左\ lfloor \ FRAC {C * J-B + C-1} {} \右\ rfloor] \\＆= \ sum_ {J = 0} ^ {M-1}（Nは\ \ lfloor \ FRAC {C * J-B +を残しC-1} {} \右\ rfloor）\\＆= N * MF（M-1、C、-b + C-1）\端{ALIGN *} $$
$ケース\ 3 $：当$ A <C \ \＆\＆\ B <C \ \＆\＆\ A = 0 $时$$ F（N、A、B、C）= 0 $$
记$ A（J）= \ \ lfloorを左\ FRAC {C * J-B + C-1} {} \右\ rfloor $
プロセスにおいて上記導出と同様の工程については省略する
要求$ G（N、A、B、C）$
$ケース\ 1 $：当$ A \ geqslant C \ || \ B \ geqslant C $时$$ \開始{ALIGN *} G（N、A、B、C）＆= \ sum_ {I = 0} ^ NI * \左\ lfloor \ FRAC {* I + B}、{C} \右\ rfloor \\＆= \ sum_ {I = 0} ^ nは式（I * \左\ lfloor \ FRAC {（\％のC ）* I +（B \％のC）}、{C} \右\ rfloor + \左\ lfloor \ FRAC {} {C} \右\ rfloor * I ^ 2 + \左\ lfloor \ FRAC {B}、{C } \右\ rfloor * I）\\＆= \ sum_ {i = 0} ^ NI * \左\ lfloor \ FRAC {（\％のC）* I +（B \％のC）}、{C} \右\ rfloor + \左\ lfloor \ FRAC {} {C} \右\ rfloor \ sum_ {i = 0} ^ NI ^ 2 + \左\ lfloor \ FRAC {B}、{C} \右\ rfloor \ sum_ {I = 0} ^ NI \\＆= G（N、\％のC、B \％のC、C）+ \左\ lfloor \ FRAC {} {C} \右\ rfloor * \ FRAC {N *（N +1）*（2N + 1）} {6} + \左\ lfloor \ FRAC {B}、{C} \右\ rfloor * \ FRAC {N *（N + 1）} {2} \端{ALIGN * } $$
$Case\ 2$: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 $$\begin{align*}h(n, a, b, c) &= \sum_{i=0}^n(\left \lfloor \frac{a*i+b}{c} \right \rfloor)^2 \\ &=\sum_{i=0}^n \sum_{j=0}^{\left \lfloor \frac{a*i+b}{c} \right \rfloor - 1} \sum_{k=0}^{\left \lfloor \frac{a*i+b}{c} \right \rfloor - 1}1\\ &=\sum_{j=0}^{m-1}\sum_{k=0}^{m-1} \sum_{i=0}^n[j<\left \lfloor \frac{a*i+b}{c} \right \rfloor\&\&k<\left \lfloor \frac{a*i+b}{c} \right \rfloor] \\ &=\sum_{j=0}^{m-1}\sum_{k=0}^{m-1}\sum_{i=0}^n[i>A(j)\&\&i>A(k))]\\ &= \sum_{j=0}^{m-1}\sum_{k=0}^{m-1}\sum_{i=0}^n[i>A(max(j,k))]\\ &= 2\sum_{j=0}^{m-1}\sum_{k=0}^{j-1}\sum_{i=0}^n[i>A(j)]+\sum_{j=0}^{m-1}\sum_{i=0}^n[i>A(j)]\\ &= 2\sum_{j=0}^{m-1}j*(n-A(j))+\sum_{j=0}^{m-1} (n-A(j))\\ &=2n\sum_{j=0}^{m-1}j-2\sum_{j=0}^{m-1}j*A(j)+m*n-\sum_{j=0}^{m-1}A(j) \\ &=2n*\frac{(m-1)*m}{2}+m*n-2g(m-1, c,-b+c-1,a)-f(m-1,c,-b+c-1,a)\end{align*}$$
$Case\ 3$: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 = rdc(2 * x * calc1(m - 1) % mod, 2 * res.sec % mod); ret.thr = add(ret.thr, rdc(m * x % 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; }