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http://www.lydsy.com/JudgeOnline/problem.php?id=3994
好吧表示完全不会。
建议先看一下Codeforces235E。
不妨设$a\leq b$
$\sum\limits_{i=1}^{a}\sum\limits_{j=1}^{b}d(ij)$
$=\sum\limits_{gcd(i,j)=1}\left \lfloor \frac{a}{i} \right \rfloor\left \lfloor \frac{b}{j} \right \rfloor$
$=\sum\limits_{}^{}\left \lfloor \frac{a}{i} \right \rfloor \left \lfloor \frac{b}{j} \right \rfloor [gcd(i,j)==1]$
$=\sum\limits_{}^{}\left \lfloor \frac{a}{i} \right \rfloor \left \lfloor \frac{b}{j} \right \rfloor\sum\limits_{d|gcd(i,j)}\varphi (d)$
$=\sum\limits_{}^{}\left \lfloor \frac{a}{i} \right \rfloor \left \lfloor \frac{b}{j} \right \rfloor\sum\limits_{d|i,d|j}\varphi (d)$
$=\sum\limits_{1\leq d\leq a}\varphi (d)\sum\limits_{d|i}^{}\sum\limits_{d|j}^{}\left \lfloor \frac{a}{i} \right \rfloor \left \lfloor \frac{b}{j} \right \rfloor$
设$i=di‘$,$j=dj‘$，则：
$\sum\limits_{1\leq d\leq a}\varphi (d)\sum\limits_{i‘}^{}\sum\limits_{j‘}^{}\left \lfloor \frac{a}{di‘} \right \rfloor \left \lfloor \frac{b}{dj‘} \right \rfloor$
记$f(n)=\sum\limits_{i=1}^{n}\left \lfloor \frac{n}{i} \right \rfloor=\sum\limits_{i=1}^{n}d(i)$
我们先用线性筛求出$d(i)$，就可以进一步求出$f(n)$
则原式为：
$\sum\limits_{i=1}^{a}\varphi (d)f(\left \lfloor \frac{a}{d} \right \rfloor)f(\left \lfloor \frac{b}{d} \right \rfloor)$
#include<cstdio> #include<cstdlib> #include<iostream> #include<fstream> #include<algorithm> #include<cstring> #include<string> #include<cmath> #include<queue> #include<stack> #include<map> #include<utility> #include<set> #include<bitset> #include<vector> #include<functional> #include<deque> #include<cctype> #include<climits> #include<complex> //#include<bits/stdc++.h>适用于CF,UOJ，但不适用于poj using namespace std; typedef long long LL; typedef double DB; typedef pair<int,int> PII; typedef complex<DB> CP; #define mmst(a,v) memset(a,v,sizeof(a)) #define mmcy(a,b) memcpy(a,b,sizeof(a)) #define fill(a,l,r,v) fill(a+l,a+r+1,v) #define re(i,a,b) for(i=(a);i<=(b);i++) #define red(i,a,b) for(i=(a);i>=(b);i--) #define ire(i,x) for(typedef(x.begin()) i=x.begin();i!=x.end();i++) #define fi first #define se second #define m_p(a,b) make_pair(a,b) #define p_b(a) push_back(a) #define SF scanf #define PF printf #define two(k) (1<<(k)) template<class T>inline T sqr(T x){return x*x;} template<class T>inline void upmin(T &t,T tmp){if(t>tmp)t=tmp;} template<class T>inline void upmax(T &t,T tmp){if(t<tmp)t=tmp;} inline int sgn(DB x){if(abs(x)<1e-9)return 0;return(x>0)?1:-1;} const DB Pi=acos(-1.0); int gint() { int res=0;bool neg=0;char z; for(z=getchar();z!=EOF && z!=‘-‘ && !isdigit(z);z=getchar()); if(z==EOF)return 0; if(z==‘-‘){neg=1;z=getchar();} for(;z!=EOF && isdigit(z);res=res*10+z-‘0‘,z=getchar()); return (neg)?-res:res; } LL gll() { LL res=0;bool neg=0;char z; for(z=getchar();z!=EOF && z!=‘-‘ && !isdigit(z);z=getchar()); if(z==EOF)return 0; if(z==‘-‘){neg=1;z=getchar();} for(;z!=EOF && isdigit(z);res=res*10+z-‘0‘,z=getchar()); return (neg)?-res:res; } const int n=50000; LL mobi[n+100],summobi[n+100],d[n+100],f[n+100]; int cnt,prime[n+100]; int flag[n+100]; LL minpx[n+100]; void get() { int i,j; mobi[1]=1; d[1]=1; re(i,2,n) { if(!flag[i])prime[++cnt]=i,mobi[i]=-1,d[i]=2,minpx[i]=1; for(j=1;j<=cnt && i*prime[j]<=n;j++) { flag[i*prime[j]]=1; if(i%prime[j]==0) { mobi[i*prime[j]]=0; d[i*prime[j]]=d[i]/(minpx[i]+1)*(minpx[i]+2); minpx[i*prime[j]]=minpx[i]+1; } else { mobi[i*prime[j]]=-mobi[i]; d[i*prime[j]]=d[i]*2; minpx[i*prime[j]]=1; } } } } int main() { freopen("divisor.in","r",stdin); freopen("divisor.out","w",stdout); int i; get(); re(i,1,n)f[i]=f[i-1]+d[i]; re(i,1,n)summobi[i]=summobi[i-1]+mobi[i]; for(int T=gint();T;T--) { int a=gint(),b=gint();LL ans=0; if(a>b)swap(a,b); for(int d=1,last;d<=a;d=last+1) { last=min(a/(a/d),b/(b/d)); upmin(last,a); ans+=(summobi[last]-summobi[d-1])*f[a/d]*f[b/d]; } PF("%I64d\n",ans); } return 0; }
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原文地址：http://www.cnblogs.com/maijing/p/4832574.html
