类欧几里得算法

浪破
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先考虑一个简单的问题 f(a,b,c,n)=∑i=0n⌊ai+bc⌋f(a,b,c,n)=\sum_{i=0}^n\lfloor\frac{ai+b}{c}\rfloor

我们这样来解决

f(a,b,c,n)=∑i=0n⌊ai+bc⌋=f(a%c,b%c,c,n)+∑i=0n(i⌊ac⌋+⌊bc⌋)=f(a%c,b%c,c,n)+n(n+1)2⌊ac⌋+(n+1)⌊bc⌋令m=⌊an+bc⌋则f(a,b,c,n)=∑i=0n⌊ai+bc⌋=∑i=0n∑j=1m[⌊ai+bc⌋≥j]=∑i=0n∑j=0m−1[⌊ai+bc⌋≥j+1]=∑i=0n∑j=0m−1[ai+b≥cj+c]=∑i=0n∑j=0m−1[ai>cj+c−b−1]=∑i=0n∑j=0m−1[i>⌊cj+c−b−1a⌋]=∑j=0m−1n−⌊cj+c−b−1a⌋=nm−f(c,c−b−1,a,m−1)可以开始递归，递归出口m=0\begin{aligned} \\&f(a,b,c,n) \\&=\sum_{i=0}^n\lfloor\frac{ai+b}{c}\rfloor \\&=f(a\%c,b\%c,c,n)+\sum_{i=0}^{n}(i\lfloor\frac{a}{c}\rfloor+\lfloor\frac{b}{c}\rfloor) \\&=f(a\%c,b\%c,c,n)+\frac{n(n+1)}{2}\lfloor\frac{a}{c}\rfloor+(n+1)\lfloor\frac{b}{c}\rfloor \\ \\&令m=\lfloor\frac{an+b}{c}\rfloor \\&则f(a,b,c,n) \\&=\sum_{i=0}^n\lfloor\frac{ai+b}{c}\rfloor \\&=\sum_{i=0}^n\sum_{j=1}^m[\lfloor\frac{ai+b}{c}\rfloor\geq j] \\&=\sum_{i=0}^n\sum_{j=0}^{m-1}[\lfloor\frac{ai+b}{c}\rfloor\geq j+1] \\&=\sum_{i=0}^n\sum_{j=0}^{m-1}[ai+b \geq cj+c] \\&=\sum_{i=0}^n\sum_{j=0}^{m-1}[ai \gt cj+c-b-1] \\&=\sum_{i=0}^n\sum_{j=0}^{m-1}[i \gt \lfloor\frac{cj+c-b-1}{a}\rfloor] \\&=\sum_{j=0}^{m-1}n-\lfloor\frac{cj+c-b-1}{a}\rfloor \\&=nm-f(c,c-b-1,a,m-1) \\&可以开始递归，递归出口 m=0 \end{aligned}

然后我们考虑两个难一点的题目，同时解决这两个问题

h(a,b,c,n)=∑i=0n⌊ai+bc⌋2g(a,b,c,n)=∑i=0ni⌊ai+bc⌋\begin{aligned} &h(a,b,c,n)=\sum_{i=0}^n\lfloor\frac{ai+b}{c}\rfloor^2 \quad\quad\quad\quad g(a,b,c,n)=\sum_{i=0}^ni\lfloor\frac{ai+b}{c}\rfloor \end{aligned}

先来看h

h(a,b,c,n)=∑i=0n⌊ai+bc⌋2=∑i=0n(⌊(a%c)i+(b%c)c⌋+⌊ac⌋i+⌊bc⌋)2=∑i=0n(⌊(a%c)i+(b%c)c⌋2+⌊ac⌋2i2+⌊bc⌋2+2⌊ac⌋i⌊bc⌋+2⌊(a%c)i+(b%c)c⌋⌊ac⌋i+2⌊(a%c)i+(b%c)c⌋⌊bc⌋=h(a%c,b%c,c,n)+2⌊ac⌋g(a%c,b%c,c,n)+2⌊bc⌋f(a%c,b%c,c,n)+⌊ac⌋2n(n+1)(2n+1)6+2⌊ac⌋⌊bc⌋n(n+1)2+(n+1)⌊bc⌋2\begin{aligned} &h(a,b,c,n)\\ =&\sum_{i=0}^n\lfloor\frac{ai+b}{c}\rfloor^2\\ =&\sum_{i=0}^n(\lfloor\frac{(a\%c)i+(b\%c) }{c}\rfloor+\lfloor\frac{a}{c}\rfloor i+\lfloor\frac{b}{c}\rfloor)^2\\ =&\sum_{i=0}^n(\lfloor\frac{(a\%c)i+(b\%c) }{c}\rfloor^2+\lfloor\frac{a}{c}\rfloor^2i^2+\lfloor\frac{b}{c}\rfloor^2+2\lfloor\frac{a}{c}\rfloor i\lfloor\frac{b}{c}\rfloor+2\lfloor\frac{(a\%c)i+(b\%c) }{c}\rfloor\lfloor\frac{a}{c}\rfloor i+2\lfloor\frac{(a\%c)i+(b\%c) }{c}\rfloor\lfloor\frac{b}{c}\rfloor\\ =&h(a\%c,b\%c,c,n)+2\lfloor\frac{a}{c}\rfloor g(a\%c,b\%c,c,n)+2\lfloor\frac{b}{c}\rfloor f(a\%c,b\%c,c,n)+\lfloor\frac{a}{c}\rfloor^2\frac{n(n+1)(2n+1)}{6}+2\lfloor\frac{a}{c}\rfloor \lfloor\frac{b}{c}\rfloor\frac{n(n+1)}{2}+(n+1)\lfloor\frac{b}{c}\rfloor^2 \end{aligned}

这里我们只用关心第一项

令m=⌊an+bc⌋则h(a,b,c,n)=∑i=0n⌊ai+bc⌋2=∑i=0n(∑j=1m[⌊ai+bc⌋≥j])2=∑i=0n(∑j=0m−1[i>⌊cj+c−b−1a⌋])2=∑i=0n∑j=0m−1[i>⌊cj+c−b−1a⌋]∑k=0m−1[i>⌊ck+c−b−1a⌋]=∑i=0n∑j=0m−1∑k=0m−1[i>⌊cj+c−b−1a⌋]∗[i>⌊ck+c−b−1a⌋]=∑i=0n∑j=0m−1∑k=0m−1[i>max(⌊cj+c−b−1a⌋,⌊ck+c−b−1a⌋)]=∑i=0n∑j=0m−1∑k=0m−1[i>max(⌊cj+c−b−1a⌋,⌊ck+c−b−1a⌋)]=nm2−∑j=0m−1∑k=0m−1max(⌊cj+c−b−1a⌋,⌊ck+c−b−1a⌋)=nm2−2∑j=0m−1j⌊cj+c−b−1a⌋−∑j=0m−1⌊cj+c−b−1a⌋=nm2−2g(c,c−b−1,a,m−1)−f(c,c−b−1,a,m−1)\begin{aligned} &令m=\lfloor\frac{an+b}{c}\rfloor则\\ &h(a,b,c,n)\\ =&\sum_{i=0}^n\lfloor\frac{ai+b}{c}\rfloor^2\\ =&\sum_{i=0}^n(\sum_{j=1}^m[\lfloor\frac{ai+b}{c}\rfloor\geq j])^2\\ =&\sum_{i=0}^n(\sum_{j=0}^{m-1}[i \gt \lfloor\frac{cj+c-b-1}{a}\rfloor])^2\\ =&\sum_{i=0}^n\sum_{j=0}^{m-1}[i \gt \lfloor\frac{cj+c-b-1}{a}\rfloor]\sum_{k=0}^{m-1}[i \gt \lfloor\frac{ck+c-b-1}{a}\rfloor]\\ =&\sum_{i=0}^n\sum_{j=0}^{m-1}\sum_{k=0}^{m-1}[i \gt \lfloor\frac{cj+c-b-1}{a}\rfloor]*[i \gt \lfloor\frac{ck+c-b-1}{a}\rfloor]\\ =&\sum_{i=0}^n\sum_{j=0}^{m-1}\sum_{k=0}^{m-1}[i \gt max(\lfloor\frac{cj+c-b-1}{a}\rfloor,\lfloor\frac{ck+c-b-1}{a}\rfloor)]\\ =&\sum_{i=0}^n\sum_{j=0}^{m-1}\sum_{k=0}^{m-1}[i \gt max(\lfloor\frac{cj+c-b-1}{a}\rfloor,\lfloor\frac{ck+c-b-1}{a}\rfloor)]\\ =&nm^2-\sum_{j=0}^{m-1}\sum_{k=0}^{m-1} max(\lfloor\frac{cj+c-b-1}{a}\rfloor,\lfloor\frac{ck+c-b-1}{a}\rfloor)\\ =&nm^2-2\sum_{j=0}^{m-1}j\lfloor\frac{cj+c-b-1}{a}\rfloor-\sum_{j=0}^{m-1}\lfloor\frac{cj+c-b-1}{a}\rfloor\\ =&nm^2-2g(c,c-b-1,a,m-1)-f(c,c-b-1,a,m-1) \end{aligned}

推出来了。。。。。 然后我们来怼g

g(a,b,c,n)=∑i=0ni⌊ai+bc⌋=∑i=0ni⌊(a%c)i+b%cc+⌊ac⌋i+⌊bc⌋⌋=∑i=0ni(⌊(a%c)i+b%cc⌋+⌊ac⌋i+⌊bc⌋)=∑i=0ni⌊(a%c)i+b%cc⌋+∑i=0n⌊ac⌋i2+∑i=0n⌊bc⌋i=n(n+1)(2n+1)6⌊ac⌋+n(n+1)2⌊bc⌋+∑i=0ni⌊(a%c)i+b%cc⌋=g(a%c,b%c,c,n)+n(n+1)(2n+1)6⌊ac⌋+n(n+1)2⌊bc⌋\begin{aligned} &g(a,b,c,n)\\ =&\sum_{i=0}^ni\lfloor\frac{ai+b}{c}\rfloor\\ =&\sum_{i=0}^ni\lfloor\frac{(a\%c)i+b\%c}{c}+\lfloor\frac{a}{c}\rfloor i+\lfloor\frac{b}{c}\rfloor\rfloor \\ =&\sum_{i=0}^ni(\lfloor\frac{(a\%c)i+b\%c}{c}\rfloor+\lfloor\frac{a}{c}\rfloor i+\lfloor\frac{b}{c}\rfloor)\\ =&\sum_{i=0}^ni\lfloor\frac{(a\%c)i+b\%c}{c}\rfloor+\sum_{i=0}^n\lfloor\frac{a}{c}\rfloor i^2+\sum_{i=0}^n\lfloor\frac{b}{c}\rfloor i\\ =&\frac{n(n+1)(2n+1)}{6}\lfloor\frac{a}{c}\rfloor +\frac{n(n+1)}{2}\lfloor\frac{b}{c}\rfloor +\sum_{i=0}^ni\lfloor\frac{(a\%c)i+b\%c}{c}\rfloor\\ =&g(a\%c,b\%c,c,n)+\frac{n(n+1)(2n+1)}{6}\lfloor\frac{a}{c}\rfloor +\frac{n(n+1)}{2}\lfloor\frac{b}{c}\rfloor \end{aligned}

同理我们只关心第一项

g(a,b,c,n)=∑i=0ni⌊ai+bc⌋=∑i=0n(i∑j=1m[⌊ai+bc⌋≥j])=∑i=0n(i∑j=0m−1[i>⌊cj+c−b−1a⌋])=∑i=0n∑j=0m−1i[i>⌊cj+c−b−1a⌋]=∑j=0m−1∑i=⌊cj+c−b−1a⌋+1ni=∑j=0m−1(n+⌊cj+c−b−1a⌋+1)∗(n−(⌊cj+c−b−1a⌋+1)+1)2=∑j=0m−1(n+⌊cj+c−b−1a⌋+1)∗(n−⌊cj+c−b−1a⌋)2=∑j=0m−1n2−⌊cj+c−b−1a⌋2+n−⌊cj+c−b−1a⌋2=n(n+1)m2−∑j=0m−1⌊cj+c−b−1a⌋22−∑j=0m−1⌊cj+c−b−1a⌋2=n(n+1)m2−h(c,c−b−1,a,m−1)2−f(c,c−b−1,a,m−1)2\begin{aligned} &g(a,b,c,n)\\ =&\sum_{i=0}^ni\lfloor\frac{ai+b}{c}\rfloor\\ =&\sum_{i=0}^n(i\sum_{j=1}^m[\lfloor\frac{ai+b}{c}\rfloor \geq j])\\ =&\sum_{i=0}^n(i\sum_{j=0}^{m-1}[i \gt \lfloor\frac{cj+c-b-1}{a}\rfloor])\\ =&\sum_{i=0}^n\sum_{j=0}^{m-1}i[i \gt \lfloor\frac{cj+c-b-1}{a}\rfloor]\\ =&\sum_{j=0}^{m-1}\sum_{i=\lfloor\frac{cj+c-b-1}{a}\rfloor+1}^ni\\ =&\sum_{j=0}^{m-1}\frac{(n+\lfloor\frac{cj+c-b-1}{a}\rfloor+1)*(n-(\lfloor\frac{cj+c-b-1}{a}\rfloor+1)+1)}{2}\\ =&\sum_{j=0}^{m-1}\frac{(n+\lfloor\frac{cj+c-b-1}{a}\rfloor+1)*(n-\lfloor\frac{cj+c-b-1}{a}\rfloor)}{2}\\ =&\sum_{j=0}^{m-1}\frac{n^2-\lfloor\frac{cj+c-b-1}{a}\rfloor^2+n-\lfloor\frac{cj+c-b-1}{a}\rfloor}{2}\\ =&\frac{n(n+1)m}{2}-\frac{\sum_{j=0}^{m-1}\lfloor\frac{cj+c-b-1}{a}\rfloor^2}{2}-\frac{\sum_{j=0}^{m-1}\lfloor\frac{cj+c-b-1}{a}\rfloor}{2}\\ =&\frac{n(n+1)m}{2}-\frac{h(c,c-b-1,a,m-1)}{2}-\frac{f(c,c-b-1,a,m-1)}{2}\\ \end{aligned}

推完了总结一下

f(a,b,c,n)=∑i=0n⌊ai+bc⌋h(a,b,c,n)=∑i=0n⌊ai+bc⌋2g(a,b,c,n)=∑i=0ni⌊ai+bc⌋f(a,b,c,n)=f(a%c,b%c,c,n)+n(n+1)2⌊ac⌋+(n+1)⌊bc⌋h(a,b,c,n)=h(a%c,b%c,c,n)+2⌊ac⌋g(a%c,b%c,c,n)+2⌊bc⌋f(a%c,b%c,c,n)+⌊ac⌋2n(n+1)(2n+1)6+2⌊ac⌋⌊bc⌋n(n+1)2+(n+1)⌊bc⌋2g(a,b,c,n)=g(a%c,b%c,c,n)+n(n+1)(2n+1)6⌊ac⌋+n(n+1)2⌊bc⌋f(a,b,c,n)=nm−f(c,c−b−1,a,m−1)h(a,b,c,n)=nm2−2g(c,c−b−1,a,m−1)−f(c,c−b−1,a,m−1)g(a,b,c,n)=n(n+1)m2−h(c,c−b−1,a,m−1)2−f(c,c−b−1,a,m−1)2\begin{aligned} &f(a,b,c,n)=\sum_{i=0}^n\lfloor\frac{ai+b}{c}\rfloor\\ &h(a,b,c,n)=\sum_{i=0}^n\lfloor\frac{ai+b}{c}\rfloor^2 \\ & g(a,b,c,n)=\sum_{i=0}^ni\lfloor\frac{ai+b}{c}\rfloor \\ \\ &f(a,b,c,n)=f(a\%c,b\%c,c,n)+\frac{n(n+1)}{2}\lfloor\frac{a}{c}\rfloor+(n+1)\lfloor\frac{b}{c}\rfloor\\ &h(a,b,c,n)=h(a\%c,b\%c,c,n)+2\lfloor\frac{a}{c}\rfloor g(a\%c,b\%c,c,n)+2\lfloor\frac{b}{c}\rfloor f(a\%c,b\%c,c,n)+\lfloor\frac{a}{c}\rfloor^2\frac{n(n+1)(2n+1)}{6}+2\lfloor\frac{a}{c}\rfloor \lfloor\frac{b}{c}\rfloor\frac{n(n+1)}{2}+(n+1)\lfloor\frac{b}{c}\rfloor^2\\ &g(a,b,c,n)=g(a\%c,b\%c,c,n)+\frac{n(n+1)(2n+1)}{6}\lfloor\frac{a}{c}\rfloor +\frac{n(n+1)}{2}\lfloor\frac{b}{c}\rfloor\\ \\ &f(a,b,c,n)=nm-f(c,c-b-1,a,m-1)\\ &h(a,b,c,n)=nm^2-2g(c,c-b-1,a,m-1)-f(c,c-b-1,a,m-1)\\ &g(a,b,c,n)=\frac{n(n+1)m}{2}-\frac{h(c,c-b-1,a,m-1)}{2}-\frac{f(c,c-b-1,a,m-1)}{2}\\ \end{aligned}
void calfgh_baoli(ll a,ll b,ll c,ll n,ll&f,ll&g,ll&h){
    f=g=h=0;
    for(ll i=0;i<=n;i++) {
        f+=(a*i+b)/c;
        g+=i*((a*i+b)/c);
        h+=((a*i+b)/c)*((a*i+b)/c);
    }
}
// a>=0 b>=0 c>0 n>=0         -> O(lg(a,c))
void calfgh(ll a,ll b,ll c,ll n,ll&f,ll&g,ll&h){
    ll A=a/c,B=b/c,s0=n+1,s1=n*(n+1)/2,s2=n*(n+1)*(2*n+1)/6;
    f=s1*A+s0*B;
    g=s2*A+s1*B;
    h=s2*A*A+s0*B*B+2*s1*A*B-2*B*f-2*A*g;// 先减掉
    a%=c,b%=c;
    ll m=(a*n+b)/c;
    if(m!=0) {
        ll ff,gg,hh; calfgh(c,c-b-1,a,m-1,ff,gg,hh);
        f+=n*m-ff;
        g+=(n*m*(n+1)-hh-ff)/2;
        h+=n*m*m-2*gg-ff;
    }
    h+=2*B*f+2*A*g;//再加上
}
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