【数论分块】[BZOJ2956、LuoguP2260] 模积和

十年OI一场空，忘记取模见祖宗

题目：

求$$\sum_{i=1}^{n}\sum_{j=1}^{m} (n \bmod i)(m \bmod i)$$

(其中i,j不相等)

暴力拆式子：

　$$ANS=\sum_{i=1}^{n}\sum_{j=1}^{m} (n- \left \lfloor \frac{n}{i} \right \rfloor*i)(m- \left \lfloor \frac{m}{i} \right \rfloor*i)-\sum_{i=1}^{min(n,m)} (n- \left \lfloor \frac{n}{i} \right \rfloor *i)(m- \left \lfloor \frac{m}{i} \right \rfloor *i)$$

令$f(n)=\sum_{i=1}^{n} (n- \left \lfloor \frac{n}{i} \right \rfloor *i)$

令$g(n)=\sum_{i=1}^{n}(n- \left \lfloor \frac{n}{i} \right \rfloor *i)(m- \left \lfloor \frac{m}{i} \right \rfloor *i)$

不妨设n<=m

则有

$$ANS=f(n)*f(m)-g(n)$$

其中$$g(n)=\sum_{i=1}^{n} n*m-n*\sum_{i=1}^{n} \left \lfloor \frac{m}{i} \right \rfloor *i-m*\sum_{i=1}^{n} \left \lfloor \frac{n}{i} \right \rfloor *i+\sum_{i=1}^{n} \left \lfloor \frac{n}{i} \right \rfloor* \left \lfloor \frac{m}{i} \right \rfloor *i^2$$

且易有$$\sum_{i=1}^{n} i^2=\frac{n*(n+1)*(2*n+1)}{6}$$

预处理6在模19940417意义下的逆元（我用了exgcd）

然后用数论分块把上面一堆东西算一下即可

 #include<bits/stdc++.h>
 #define int long long
 #define writeln(x)  write(x),puts("")
 #define writep(x)   write(x),putchar(' ')
 using namespace std;
 inline int read(){
     int ans=,f=;char chr=getchar();
     while(!isdigit(chr)){if(chr=='-') f=-;chr=getchar();}
     while(isdigit(chr)){ans=(ans<<)+(ans<<)+chr-;chr=getchar();}
     return ans*f;
 }void write(int x){
     if(x<) putchar('-'),x=-x;
     if(x>) write(x/);
     putchar(x%+'');
 }const int mod = ;
 int n,m,k;
 inline void Add(int &x,int y){x+=y;x%=mod;}
 void exgcd(int a,int b,int &x,int &y){
     if(b==)return x=,y=,void();
     exgcd(b,a%b,x,y);
     int t=x;x=y,y=t-a/b*y;
 }int inv(int x){
     int xx,y;
     exgcd(,mod,xx,y);
     xx=(xx%mod+mod)%mod;
     return xx;
 }const int inv6=inv();
 int sum(int x){return (x)*(x+)%mod*(*x%mod+)%mod*inv6%mod;}
 int query1(int l,int r){return ((sum(r)-sum(l-))%mod+mod)%mod;}
 int query2(int l,int r){int ans=(r-l+)*(l+r)/;return ans%mod;}
 int calc1(int n){
     int ans=;
     for(int i=,j,t;i<=n;i=j+){
         j=n/(n/i);
         t=n/i*(i+j)*(j-i+)/;
         t%=mod;
         Add(ans,t);
     }ans=n*n%mod-ans;
     ans=(ans%mod+mod)%mod;
     return ans;
 }int calc2(int k){
     int ans=;
     for(int i=,j,t;i<=n;i=j+){
         j=min(n/(n/i),m/(m/i));
         int s1=n*(m/i)%mod*query2(i,j)%mod;
         int s2=m*(n/i)%mod*query2(i,j)%mod;
         int s3=(n/i)*(m/i)%mod*query1(i,j)%mod;
         Add(s1,s2);
         Add(ans,s1);
         ans-=s3;
         ans=((ans)%mod+mod)%mod;
     }return ans;
 }
 signed main(){
     n=read(),m=read();
     if(n>m)swap(n,m);
     int ans=calc1(n)*calc1(m)%mod;
     ans-=n*m%mod*n%mod;
     ans=(ans%mod+mod)%mod;
     ans+=calc2(n);
     ans=(ans%mod+mod)%mod;
     cout<<ans<<endl;
     return ;
 }