类欧几里得求解问题有如下形式： $f(a,b,c,n)={\sum }_{i=0}^{n-1}\lfloor \frac{ai+b}{c}\rfloor$

普通floor_sum

若$a\ge c$或者$b\ge c$,令$a^{\prime }=a\mathrm{mod}c,b^{\prime }=b\mathrm{mod}c$
$$f(a,b,c,n)=\sum _{i=0}^{n-1}\lfloor \frac{a}{c}\rfloor i+\lfloor \frac{b}{c}\rfloor +\lfloor \frac{a^{\prime }i+b^{\prime }}{c}\rfloor =\frac{n(n-1)}{2}\lfloor \frac{a}{c}\rfloor +n\lfloor \frac{b}{c}\rfloor +f(a^{\prime },b^{\prime },c,n)$$
就转化成了$a<c,b<c$的情况

考虑$a<c,b<c$的情况：
$$f(a,b,c,n)=\sum _{i=0}^{n-1}\sum _{j=0}^{\lfloor \frac{ai+b}{c}\rfloor -1}[1]$$
令$m=\lfloor \frac{a(n-1)+b}{c}\rfloor -1$
$$\begin{gathered} f(a,b,c,n)=\sum _{j=0}^m\sum _{i=0}^{n-1}[j<\lfloor \frac{ai+b}{c}\rfloor ] \\ =\sum _{j=0}^{m-1}\sum _{i=0}^{n-1}[c(j+1)\le ai+b] \\ =\sum _{j=0}^{m-1}\sum _{i=0}^{n-1}[i>\lfloor \frac{cj+c-b-1}{a}\rfloor ] \\ =\sum _{j=0}^{m-1}[n-1-\lfloor \frac{cj+c-b-1}{a}\rfloor ] \\ =(n-1)m-f(c,c-b-1,a,m) \end{gathered}$$
复杂度为$O(\log \max (a,c))$

从几何意义讲，表示的是$y=\frac{ax+b}{c}$下的整数点

类似地，$g(a,b,c,n)={\sum }_{i=0}^{n-1}i\lfloor \frac{ai+b}{c}\rfloor ,h(a,b,c,n)={\sum }_{i=0}^{n-1}\lfloor \frac{ai+b}{c}{\rfloor }^2$

$a\ge c,b\ge c$的情况:

$g(a,b,c,n)=\frac{n(n-1)(2n-1)}{6}\lfloor \frac{a}{c}\rfloor +\frac{n(n-1)}{2}\lfloor \frac{b}{c}\rfloor +g(a^{\prime },b^{\prime },c,n)$

$h(a,b,c,n)=h(a^{\prime },b^{\prime },c,n)+2\lfloor \frac{a}{c}\rfloor g(a^{\prime },b^{\prime },c,n)+2\lfloor \frac{b}{c}\rfloor f(a^{\prime },b^{\prime },c,n)+\frac{n(n-1)(2n-1)}{6}\lfloor \frac{a}{c}{\rfloor }^2+n(n-1)\lfloor \frac{a}{c}\rfloor \lfloor \frac{b}{c}\rfloor +n\lfloor \frac{b}{c}{\rfloor }^2$

$a<c,b<c$的情况：

$g(a,b,c,n)=\frac{1}{2}(mn(n-1)-h(c,c-b-1,a,m)-f(c,c-b-1,a,m))$

$h(a,b,c,n)=(n-1)m^2-2g(c,c-b-1,a,m)-f(c,c-b-1,a,m)$

洛谷P5170 【模板】类欧几里得算法

struct FloorSum{
    static constexpr int MOD=998244353,inv2=499122177,inv6=166374059;
    struct Node{
        int f=0,g=0,h=0;
    };
    int mo(int x){return x>=MOD?x-MOD:x;}
    int sqr(int x){return 1ll*x*x%MOD;}
    Node f(int a,int b,int c,int n){
        Node res;
        if(!a){
            res.f=1ll*b/c*n%MOD;
            res.g=1ll*n*(n-1)/2%MOD*(b/c)%MOD;
            res.h=1ll*sqr(b/c)*n%MOD;
        }
        else if(a>=c||b>=c){
            Node o=f(a%c,b%c,c,n);
            res.f=(o.f+1ll*n*(n-1)/2%MOD*(a/c)+1ll*b/c*n)%MOD;
            res.g=(o.g+1ll*a/c*n%MOD*(n-1)%MOD*(n*2-1)%MOD*inv6+1ll*n*(n-1)/2%MOD*(b/c))%MOD;
            res.h=(o.h+2ll*(a/c)*o.g+2ll*(b/c)*o.f+1ll*n*(n-1)%MOD*(n*2-1)%MOD*inv6%MOD*sqr(a/c)+1ll*n*(n-1)%MOD*(a/c)%MOD*(b/c)+1ll*n*sqr(b/c))%MOD;
        }
        else{
            int m=(1ll*a*(n-1)+b)/c;
            Node o=f(c,c-b-1,a,m);
            res.f=(1ll*(n-1)*m-o.f+MOD)%MOD;
            res.g=((1ll*m*n%MOD*(n-1)-o.h-o.f)%MOD+MOD)*inv2%MOD;
            res.h=(1ll*(n-1)*m%MOD*m-mo(o.g*2)-o.f+MOD+MOD)%MOD;
        }
        return res;
    }
};
void solve(){
    FloorSum f;
    int a,b,c,n;
    cin>>n>>a>>b>>c;
    auto res=f.f(a,b,c,n+1);
    cout<<res.f<<" "<<res.h<<" "<<res.g<<"\n";
}

万能欧几里得算法

具体内容参考这篇博客：https://www.cnblogs.com/apjifengc/p/17492207.html

下面只讲做法推导
记如果经过网格的一条横线，$y=c(c是大于等于0的任意正整数)$，就乘上矩阵$U$，如果经过$x=c(c是大于等于0的任意正整数)$，乘上矩阵$R$,如果同时经过则先$U$再$R$

例子：$y=\lfloor \frac{ai+b}{c}\rfloor$,我们维护向量$[y,\sum y,1]$

则经过$U$时，$y+1$，经过$R$，$\sum y+y$
$$U=\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 1& 0& 1\end{array}\right),R=\left(\begin{array}{ccc}1& 1& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)$$
模板如下，复杂度为$O(T\log \max (a,c))$

template<typename T>
struct FloorSum{
    T qpow(T a,int b){
        T res;
        while(b){
            if(b&1)res=res*a;
            a=a*a;
            b>>=1;
        }
        return res;
    }
    T f(int a,int b,int c,int n,T U,T R){
        if(!n)return T();
        if(b>=c)return qpow(U,b/c)*f(a,b%c,c,n,U,R);
        if(a>=c)return f(a%c,b,c,n,U,qpow(U,a/c)*R);
        int m=((__int128)a*n+b)/c;
        if(!m)return qpow(R,n);
        return qpow(R,(c-b-1)/a)*U*f(c,(c-b-1)%a,a,m-1,R,U)*qpow(R,n-((__int128)c*m-b-1)/a);
    }
};

考虑模板题，要求的是$\sum y,\sum y^2,\sum xy$，如何合并两个区间

定义$X_{l/r},Y_{l/r}$表示左/右区间的$cntr,cntu$的值(经过多少个$R，U$)
$$\begin{gathered} \sum y=\sum y_l+\sum (y_r+Y_l)=\sum y_l+\sum y_r+X_r{Y}_l \\ \sum y^2=\sum y_l^2+\sum (y_r+Y_l{)}^2=\sum y_l^2+\sum y_r^2+2Y_l\sum y_r+X_r{Y}_l^2 \\ \sum xy=\sum x_l{y}_l+\sum (x_r+X_l)(y_r+Y_l)=\sum x_l{y}_l+\sum x_r{y}_r+Y_l\sum x_r+X_l\sum y_r+X_r{X}_l{Y}_l \\ \sum x=\sum x_l+\sum (x_r+X_l)=\sum x_l+\sum x_r+X_l{X}_r \end{gathered}$$
总结：

• 左边求和的数量为$X_l$，右边求和的数量为$X_r$
• 合并时，右边的$x,y$加上$X_l,Y_l$
• $U,R$初始化时，$U$初始化$y$，$R$初始化$x$以及其他和$x$相关的答案

写出如下结构体

struct Info{
    int x,y,sumx,sumy,sumxy,sumy2;
    Info():x(0),y(0),sumx(0),sumy(0),sumxy(0),sumy2(0){}
    Info operator * (const Info &r){
        Info res;
        res.x=(x+r.x)%mod;
        res.y=(y+r.y)%mod;
        res.sumx=(sumx+r.sumx+x*r.x%mod)%mod;
        res.sumy=(sumy+r.sumy+y*r.x%mod)%mod;
        res.sumy2=(sumy2+r.sumy2+2*y%mod*r.sumy%mod+r.x*y%mod*y%mod)%mod;
        res.sumxy=(sumxy+r.sumxy+y*r.sumx%mod+x*r.sumy%mod+x*y%mod*r.x%mod)%mod;
        return res;
    }
};

初始化$U,R$，求解的是$[1,n]$的范围，需要加上$i=0$的答案

void solve(){
    int n,a,b,c;
    cin>>n>>a>>b>>c;
    Info U,R;
    U.y=R.x=R.sumx=1;
    FloorSum<Info> f;
    auto ans=f.f(a,b,c,n,U,R);
    ans.sumy=(ans.sumy+b/c)%mod;
    ans.sumy2=(ans.sumy2+(b/c)*(b/c))%mod;
    cout<<ans.sumy<<" "<<ans.sumy2<<" "<<ans.sumxy<<"\n";
}

求${\sum }_{i=1}^n{A}^i{B}^{\lfloor \frac{ai+b}{c}\rfloor }$

维护$\sum A^x{B}^y,A^X,B^Y$
$$\sum A^x{B}^y=\sum A^{x_l}{B}^{y_l}+\sum A^{X_l+x_r}{B}^{Y_l+y_r}=\sum A^{x_l}{B}^{y_l}+A^{X_l}\sum A^{x_r}{B}^{y_r}{B}^{Y_l}$$

struct Matrix{
    static constexpr int maxn=22;
    int m[maxn][maxn];
    Matrix(){
        memset(m,0,sizeof m);
    }
    Matrix(int n){
        init(n);
    }
    void init(int n){
        memset(m,0,sizeof m);
        for(int i=0;i<n;i++)m[i][i]=1;
    }
    Matrix operator*(const Matrix &a){
        Matrix res;
        for(int k=0;k<maxn;k++){
            for(int i=0;i<maxn;i++){
                for(int j=0;j<maxn;j++){
                    res.m[i][j]=(res.m[i][j]+m[i][k]*a.m[k][j]%mod)%mod;
                }
            }
        }
        return res;
    }
    Matrix operator+(const Matrix &a){
        Matrix res;
        for(int i=0;i<maxn;i++){
            for(int j=0;j<maxn;j++){
                res.m[i][j]=(m[i][j]+a.m[i][j])%mod;
            }
        }
        return res;
    }
};
struct Info{
    Matrix sum,x,y;
    Info():sum(),x(22),y(22){}
    Info operator * (const Info &r){
        Info res;
        res.x=x*r.x;
        res.y=y*r.y;
        res.sum=sum+x*r.sum*y;
        return res;
    }
};
template<typename T>
struct FloorSum{
    T qpow(T a,int b){
        T res;
        while(b){
            if(b&1)res=res*a;
            a=a*a;
            b>>=1;
        }
        return res;
    }
    T f(int a,int b,int c,int n,T U,T R){
        if(!n)return T();
        if(b>=c)return qpow(U,b/c)*f(a,b%c,c,n,U,R);
        if(a>=c)return f(a%c,b,c,n,U,qpow(U,a/c)*R);
        int m=((__int128)a*n+b)/c;
        if(!m)return qpow(R,n);
        return qpow(R,(c-b-1)/a)*U*f(c,(c-b-1)%a,a,m-1,R,U)*qpow(R,n-((__int128)c*m-b-1)/a);
    }
};
void solve(){
    int p,q,r,l,n;
    cin>>p>>q>>r>>l>>n;
    Info R,U;
    for(int i=0;i<n;i++){
        for(int j=0;j<n;j++)cin>>R.x.m[i][j],R.sum.m[i][j]=R.x.m[i][j];
    }
    for(int i=0;i<n;i++){
        for(int j=0;j<n;j++)cin>>U.y.m[i][j];
    }
    FloorSum<Info> f;
    auto ans=f.f(p,r,q,l,U,R);
    for(int i=0;i<n;i++){
        for(int j=0;j<n;j++)cout<<ans.sum.m[i][j]<<" \n"[j==n-1];
    }
}

求${\sum }_{i=0}^n\lfloor \frac{ai+b}{c}\rfloor$(简单版)

int FloorSum(int a,int b,int c,int n){// \sum_{i=0}^{n} (ai+b)/c
	if(!a)return b/c*(n+1);
	if(a<0){
		b+=n*a;
		a=-a;
	}
	if(b<0){
		int d=(a-b-1)/a;
		n-=d;
		b+=a*d;
        if(n<0)return 0;
    }
    //======== 负数计入答案 ======================
    // if(b<0){
    //     int k=(c-b-1)/c;
    //     return FloorSum(a,b+k*c,c,n)-k*(n+1);
	// }
    //===========================================
	if(a>=c||b>=c)return n*(n+1)/2*(a/c)+(b/c)*(n+1)+FloorSum(a%c,b%c,c,n);
	int m=(a*n+b)/c;
	return m*n-FloorSum(c,c-b-1,a,m-1);
}

求1~n除M模R的数的二进制1的数量

结论：对于一个数$A$二进制第$k$位为$1$，有$\lfloor \frac{A+2^k}{2^{k+1}}\rfloor -\lfloor \frac{A}{2^{k+1}}\rfloor =1$，另一种写法是$\lfloor \frac{A}{2^k}\rfloor -2\lfloor \frac{A}{2^{k+1}}\rfloor =1$

可以用类欧求解

求${\sum }_{x=0}^n{x}^{k_1}\lfloor \frac{ax+b}{c}{\rfloor }^{k_2}$

$$\begin{gathered} \sum x^{k_1}{y}^{k_2}=\sum x_l^{k_1}{y}_l^{k_2}+\sum (x_r+X_l{)}^{k_1}(y_r+Y_l{)}^{k_2} \\ \sum (x_r+X_l{)}^{k_1}(y_r+Y_l{)}^{k_2}=\sum \sum _{i=0}^{k_1}\sum _{j=0}^{k_2}\left(\genfrac{}{}{0ex}{}{k_1}{i}\right)X_l^i{x}_r^{k_1-i}\left(\genfrac{}{}{0ex}{}{k_2}{j}\right)Y_l^j{y}_r^{k_2-j}=\sum _{i=0}^{k_1}\sum _{j=0}^{k_2}\left(\genfrac{}{}{0ex}{}{k_1}{i}\right)\left(\genfrac{}{}{0ex}{}{k_2}{j}\right)X_l^i{Y}_l^j\sum x_r^{k_1-i}{y}_r^{k_2-j} \end{gathered}$$

因此启发我们维护$x,y,\sum x^i{y}^j$

struct Info{
    int x,y;
    vector<vector<int>> sum;
    Info():sum(vector (k1+1,vector<int>(k2+1))),x(0),y(0){}
    Info operator * (const Info &r){
        Info res;
        res.x=(x+r.x)%mod;
        res.y=(y+r.y)%mod;
        int cx=x%mod,cy=y%mod;
        for(int u=0;u<=k1;u++){
            for(int v=0;v<=k2;v++){
                res.sum[u][v]=sum[u][v];
                for(int i=0,X=1;i<=u;i++,X=X*cx%mod){
                    for(int j=0,Y=1;j<=v;j++,Y=Y*cy%mod){
                        res.sum[u][v]=(res.sum[u][v]+comb.binom(u,i)*comb.binom(v,j)%mod*X%mod*Y%mod*r.sum[u-i][v-j]%mod)%mod;
                    }
                }
            }
        }
        return res;
    }
};
void solve(){
    int n,a,b,c;
    cin>>n>>a>>b>>c>>k1>>k2;
    FloorSum<Info> f;
    Info U,R;
    U.y=R.x=1;
    for(int i=0;i<=k1;i++)R.sum[i][0]=1;
    comb.init(15);
    auto ans=f.f(a,b,c,n,U,R);
    int res=ans.sum[k1][k2];
    if(k1==0)res=(res+comb.qpow(b/c,k2))%mod;//k1=0，0^0=1,答案记得加上
    cout<<res<<"\n";
}

求$ax+by\le n$的对数$(x,y)$

${\sum }_{i=0}^{\frac{n}{a}}\lfloor \frac{n-ia+b}{b}\rfloor$

void solve(){
	int n,a,b;
	cin>>n>>a>>b;
	cout<<FloorSum(-a,n+b,b,n/a)<<'\n';
}

其他例题

CF1098E

给定长度为$n$的序列$a_i$，计算出每个$\gcd (a_l,....a_r)$放进$b$数组，从大到小排序；再计算$\sum b[l,...r]$放进数组$c$，然后求出数组$c$的中位数，$1\le n\le 5\ast 10^4,1\le a_i\le 10^5$

思路：
首先计算$b$数组，以$i$为右端点，向左gcd最多迭代$\log$次，st表预处理区间gcd，然后二分跳$\log$次，处理出$b$每个数字的出现次数，这部分为$O(n{\log }^{3}n)$
接下来二分答案mid，然后是推式子

• 如果区间都是相同的数，计算出最多容纳的数字mx=min(cnt[i],mid/nums[i]),然后就是等差数列求和ans+=mx*(2*cnt[i]-mx+1)/2;
• 然后是有不同的数字的情况，假设当前区间为$[l,r]$，那么记$sum=b[l+1,.....r-1]$，记$A=num{s}_l,SA=cn{t}_l,B=num{s}_r,SB=cn{t}_r$
• 如果$A\ast SA+sum+B\ast SB\le mid$那么$SA$个左端点和$SB$个右端点任选，答案为$SA\ast SB$
• 否则答案为${\sum }_{i=1}^{SA}{\sum }_{j=1}^{SB}sum+Ai+Bj\le mid$

$$\begin{gathered} j\le \frac{mid-sum-Ai}{B} \\ \sum _{i=1}^{SA}\sum _{j=1}^{SB}sum+Ai+Bj\le mid=\sum _{i=1}^{SA}\min (\max (0,\frac{mid-sum-Ai}{B}),SB) \\ \text{因为}\min (a,b)=a-max(0,a-b) \\ =\sum _{i=1}^{SA}\max (0,\frac{mid-sum-Ai}{B})-\sum _{i=1}^{SA}\max (0,\max (0,\frac{mid-sum-Ai}{B})-SB)= \\ \sum _{i=1}^{SA}\max (0,\frac{mid-sum-Ai}{B})-\sum _{i=1}^{SA}\max (0,\max (-SB,\frac{mid-sum-Ai-B\ast SB}{B})) \\ =\sum _{i=1}^{SA}\max (0,\frac{mid-sum-Ai}{B})-\sum _{i=1}^{SA}\max (0,\frac{mid-sum-Ai-B\ast SB}{B}) \end{gathered}$$
然后就可以套板子了
可以双指针维护左右端点

int FloorSum(int a,int b,int c,int n){// \sum_{i=0}^{n} (ai+b)/c
	if(!a)return b/c*(n+1);
	if(a<0){
		b+=n*a;
		a=-a;
	}
	if(b<0){
		int d=(a-b-1)/a;
		n-=d;
		b+=a*d;
	}
	if(n<0)return 0;
	if(a>=c||b>=c)return n*(n+1)/2*(a/c)+(b/c)*(n+1)+FloorSum(a%c,b%c,c,n);
	int m=(a*n+b)/c;
	return m*n-FloorSum(c,c-b-1,a,m-1);
}
template<typename T,typename F>
struct SpraseTable{
    int n;
    vector<vector<T>> st;
    F op;
    void init(int n,const vector<T>& a){
        this->n=n;
        st.assign(n+1,vector<T>(__lg(n)+1));
        for(int i=1;i<=n;i++)st[i][0]=a[i];
        for(int j=1;1<<j<=n;j++){
            for(int i=1;i+(1<<j)-1<=n;i++){
                st[i][j]=op(st[i][j-1],st[i+(1<<j-1)][j-1]);
            }
        }
    }
    T query(int l,int r){
        int k=__lg(r-l+1);
        return op(st[l][k],st[r-(1<<k)+1][k]);
    }
};
SpraseTable<int,decltype([](int a,int b){return __gcd(a,b);})> st;
void solve(){
	int n;
	cin>>n;
	vector<int> a(n+1);
	for(int i=1;i<=n;i++)cin>>a[i];
	st.init(n,a);
	map<int,int> mp;
	for(int i=1;i<=n;i++){
		int x=i;
		while(x>=1){
			int l=1,r=x;
			int g=st.query(x,i);
			while(l<r){
				int mid=l+r>>1;
				if(st.query(mid,i)==g)r=mid;
				else l=mid+1;
			}
			mp[g]+=x-l+1;
			x=l-1;
		}
	}
	vector<int> nums,cnt;
	for(auto &[x,y]:mp){
		nums.push_back(x);
		cnt.push_back(y);
	}
	int C=n*(n+1)/2;
	C=C*(C+1)/2;
	if(C&1)C=(C+1)/2;
	else C=C/2;
	auto check=[&](int mid){
		int ans=0;
		for(int i=0;i<nums.size();i++){
			int mx=min(cnt[i],mid/nums[i]);
			ans+=mx*(2*cnt[i]-mx+1)/2;
		}
		if(nums.size()<=1)return ans>=C;
		auto get=[&](int l,int r,int sum){
			if(r>=nums.size())return 0ll;
			int a=nums[l],b=nums[r],sa=cnt[l],sb=cnt[r];
			if(a*sa+sum+b*sb<=mid)return sa*sb;
			return FloorSum(-a,mid-sum,b,sa)-max(0ll,(mid-sum)/b)-FloorSum(-a,mid-sum-b*sb,b,sa-1)+max(0ll,(mid-sum-b*sb)/b);
		};
		for(int i=0,j=1,k=0,sum=0,tot=0;i<nums.size();i++){
			if(nums[i]>mid)break;
			if(i>=j)j=i+1,sum=tot=0;
			ans+=tot*cnt[i];
			while(j<nums.size()&&sum+cnt[j]*nums[j]<=mid){
				ans+=get(i,j,sum);
				sum+=cnt[j]*nums[j];
				tot+=cnt[j];
				j++;
			}
			if(j<nums.size())ans+=get(i,j,sum);
			if(i+1<j){
				sum-=cnt[i+1]*nums[i+1];
				tot-=cnt[i+1];
			}
		}
		return ans>=C;
	};
	int l=1,r=1e15;
	while(l<r){
		int mid=l+r>>1;
		if(check(mid))r=mid;
		else l=mid+1;
	}
	cout<<l<<"\n";
}

CF1182F

题意：
求$[a,b]$内$f(x)=|\sin \frac{p}{q}\pi x|$的最大值对应的最小整数$x$，$0\le a,b\le 10^9,1\le p,q\le 10^9$

思路：

• $\frac{px\mathrm{mod}q}{q}$接近$\frac{1}{2}$更优，二分这个距离，即$2px\mathrm{mod}2q\in [q-mid,q+mid]$
• 结论:$[px\mathrm{mod}q\in [l,r]]=[\lfloor \frac{px-l}{q}\rfloor -\lfloor \frac{px-r-1}{q}\rfloor >0]$
• 因此只需求和判断是否存在即可
• 求出最小距离后，用exgcd还原求解

int FloorSum(int a,int b,int c,int n){// \sum_{i=0}^{n} (ai+b)/c
	if(!a)return b/c*(n+1);
	if(a<0){
		b+=n*a;
		a=-a;
	}
	// if(b<0){
	// 	int d=(a-b-1)/a;
	// 	n-=d;
	// 	b+=a*d;
    //     if(n<0)return 0;
    // }
    //======== 负数计入答案 ======================
    if(b<0){
        int k=(c-b-1)/c;
        return FloorSum(a,b+k*c,c,n)-k*(n+1);
	}
    //===========================================
	if(a>=c||b>=c)return n*(n+1)/2*(a/c)+(b/c)*(n+1)+FloorSum(a%c,b%c,c,n);
	int m=(a*n+b)/c;
	return m*n-FloorSum(c,c-b-1,a,m-1);
}
int exgcd(int a,int b,int &x,int &y){
	if(!b){
		x=1,y=0;
		return a;
	}
	int g=exgcd(b,a%b,y,x);
	y-=a/b*x;
	return g;
}
void solve(){
	int a,b,p,q;
	cin>>a>>b>>p>>q;
	int P=p<<1,Q=q<<1;
	auto check=[&](int mid){
		int l=q-mid,r=q+mid;
		return FloorSum(P,-l,Q,b)-FloorSum(P,-l,Q,a-1)-FloorSum(P,-r-1,Q,b)+FloorSum(P,-r-1,Q,a-1)>0;
	};
	int l=0,r=1e9;
	while(l<r){
		int mid=l+r>>1;
		if(check(mid))r=mid;
		else l=mid+1;
	}
	int x,y;
	int g=exgcd(P,Q,x,y);
	int ans=1e9;
	int dx=Q/g,dy=P/g;
	if((q-l)%g==0){
		int val=(q-l)/g*x;
		val+=(a-val)/dx*dx;
		while(val>=a)val-=dx;
		while(val<a)val+=dx;
		ans=min(ans,val);
	}
	if((q+l)%g==0){
		int val=(q+l)/g*x;
		val+=(a-val)/dx*dx;
		while(val>=a)val-=dx;
		while(val<a)val+=dx;
		ans=min(ans,val);
	}
	cout<<ans<<"\n";
}

abc313G

abc402G

2020ICPC沈阳I

$$\begin{gathered} \sum _{i=0}^{H-1}\sum _{j=0}^{M-1}[|\frac{2\pi (iM+j)}{HM}-\frac{2\pi j}{M}|\le \frac{2\pi A}{HM}] \\ \sum _{i=0}^{H-1}\sum _{j=0}^{M-1}[|iM+j-Hj|\le A] \\ HM-(\sum _{i=0}^{H-1}\sum _{j=0}^{M-1}[iM+j-Hj>A]+\sum _{i=0}^{H-1}\sum _{j=0}^{M-1}[iM+j-Hj<-A]) \end{gathered}$$

对于前半部分：
$$\begin{gathered} \sum _{i=0}^{H-1}\sum _{j=0}^{M-1}[iM+j-Hj>A]= \\ \sum _{i=0}^{H-1}\sum _{j=0}^{M-1}[i>\lfloor \frac{A+(H-1)j}{M}\rfloor ]= \\ \sum _{j=0}^{M-1}H-1-\lfloor \frac{A+(H-1)j}{M}\rfloor = \\ M(H-1)-\sum _{j=0}^{M-1}\lfloor \frac{A+(H-1)j}{M}\rfloor \end{gathered}$$
对于后半部分：
$$\begin{gathered} \sum _{i=0}^{H-1}\sum _{j=0}^{M-1}[iM+j-Hj<-A]= \\ \sum _{i=0}^{H-1}\sum _{j=0}^{M-1}[i<\lfloor \frac{(H-1)j-A}{M}\rfloor ]= \\ \sum _{i=0}^{H-1}\sum _{j=0}^{M-1}[i\le \lfloor \frac{(H-1)j-A-1}{M}\rfloor ]= \\ \sum _{j=0}^{M-1}\lfloor \frac{(H-1)j-A-1}{M}\rfloor +1= \\ \sum _{j=0}^{M-1}\lfloor \frac{(H-1)j-A-1+M}{M}\rfloor \end{gathered}$$
似乎要计算负数部分和特判HM

int FloorSum(int a,int b,int c,int n){// \sum_{i=0}^{n} (ai+b)/c
	if(!a)return b/c*(n+1);
	if(a<0){
		b+=n*a;
		a=-a;
	}
	// if(b<0){
	// 	int d=(a-b-1)/a;
	// 	n-=d;
	// 	b+=a*d;
    //     if(n<0)return 0;
    // }
    //======== 负数计入答案 ======================
    if(b<0){
        int k=(c-b-1)/c;
        return FloorSum(a,b+k*c,c,n)-k*(n+1);
	}
    //===========================================
	if(a>=c||b>=c)return n*(n+1)/2*(a/c)+(b/c)*(n+1)+FloorSum(a%c,b%c,c,n);
	int m=(a*n+b)/c;
	return m*n-FloorSum(c,c-b-1,a,m-1);
}
void solve(){
    int H,M,A;
    cin>>H>>M>>A;
    int ans=H*M;
    ans-=M*(H-1)-FloorSum(H-1,A,M,M-1);
    ans-=FloorSum(H-1,M-A-1,M,M-1);
    cout<<min(H*M,ans)<<"\n";
}

补一发易理解的正解，设时间为$T$，时针角速度${\omega }_1=\frac{2\pi }{HM}$，分针角速度${\omega }_2=\frac{2\pi }{M}$

因此$\theta =T\frac{2\pi (H-1)}{HM}\mathrm{mod}2\pi$,$\min (\theta ,2\pi -\theta )\le \frac{2\pi A}{HM}$

答案就是$[T(H-1)\mathrm{mod}HM\le A]+[T(H-1)\mathrm{mod}HM\ge HM-A]$

当$A=\frac{HM}{2}$时会有重合，此时相当于角度小于等于$\pi$，则任意时刻都满足，答案为$HM$

否则来看一下式子$T(H-1)\mathrm{mod}HM\le A$，令$G=\gcd (H-1,HM)$

如果$G=1$，因为$T\in [0,HM-1]$是$HM$的完全剩余系，所以$T(H-1)$也是$HM$的完全剩余系，答案就是$A+1$

否则，考虑$T\frac{H-1}{G}\mathrm{mod}\frac{HM}{G}\le \lfloor \frac{A}{G}\rfloor$，变成上面的情况，答案就是$\lfloor \frac{A}{G}\rfloor +1$，相当于将$[0,HM-1]$切分成$G$段，因此总答案为$G(\lfloor \frac{A}{G}\rfloor +1)$

考虑$T(H-1)\mathrm{mod}HM\ge HM-A$

$T\frac{H-1}{G}\mathrm{mod}\frac{HM}{G}\ge \lceil \frac{HM-A}{G}\rceil$,答案就是$G(\frac{HM-1}{G}-\lfloor \frac{HM-A+G-1}{G}\rfloor +1)$

void solve(){
    int H,M,A;
    cin>>H>>M>>A;
    if(2*A==H*M){
        cout<<H*M<<"\n";
        return;
    }
    int G=__gcd(H-1,H*M);
    cout<<G*(A/G+1+(H*M-1)/G-(H*M-A+G-1)/G+1)<<"\n";
}

LOJ6344

按位考虑一个数第$k$位为$1$，等价于$\lfloor \frac{x}{2^k}\rfloor -2\lfloor \frac{x}{2^{k+1}}\rfloor$

定义$an{s}_i$表示${\sum }_{i=1}^n\lfloor \frac{n-i\lfloor \frac{n}{i}\rfloor }{2^k}\rfloor$

对于小于等于$\sqrt{n}$的数暴力做，这些块比较密集；大于的用整除分块+类欧

int FloorSum(int a, int b, int c, int n) { // \sum_{i=0}^{n} (ai+b)/c
    if (!a)
        return b / c * (n + 1);

    if (a < 0) {
        b += n * a;
        a = -a;
    }

    if (b < 0) {
        int d = (a - b - 1) / a;
        n -= d;
        b += a * d;

        if (n < 0)
            return 0;
    }

    //======== 负数计入答案 ======================
    // if(b<0){
    //     int k=(c-b-1)/c;
    //     return FloorSum(a,b+k*c,c,n)-k*(n+1);
    // }
    //===========================================
    if (a >= c || b >= c)
        return n * (n + 1) / 2 * (a / c) + (b / c) * (n + 1) + FloorSum(a % c, b % c, c, n);

    int m = (a * n + b) / c;
    return m * n - FloorSum(c, c - b - 1, a, m - 1);
}
void solve() {
    int n;
    cin >> n;
    int c = 0;

    for (int i = 1; i <= min(n, (int)1e6); i++)
        c ^= n % i;

    vector<int> ans(50);

    for (int l = 1e6 + 1, r; l <= n; l = r + 1) {
        int v = n / l;
        r = n / (n / l);

        for (int i = 0; i < 40; i++) {
            ans[i] += FloorSum(-v, n - l * v, 1ll << i, r - l);
        }
    }

    int res = 0;

    for (int i = 0; i < 39; i++) {
        res += ((ans[i] - 2 * ans[i + 1]) & 1) * (1ll << i);
    }

    cout << (c ^ res) << "\n";
}