bzoj3994:[SDOI2015]约数个数和

题目链接

bzoj 3994: [SDOI2015]约数个数和

题解

结论与结论的证明参考了rqy的博客
计算\(d(ij)\)时,把ij的每个约数d映射到\((a=gcd(d,i),b=(gcd(d,i),\frac{d}{gcd(d, i)})\)
可知\(a,b\)分别是\(i,j\)的因数,且\((a,b)\)对应一个因数当且仅当\(gcd(\frac ia, b) = 1\),所以

\[d(ij) = \sum_{x|i}\sum_{y|j} [gcd(\frac ix, y) = 1] = \sum_{x'|i}\sum_{y|j} [gcd(x', y) = 1] \]

有了这个式子之后

\[\sum_{i=1}^n\sum_{j=1}^m d(ij) =\sum_{i=1}^n\sum_{j=1}^m\sum_{x|i}\sum_{y|j} [gcd(x, y) = 1]$$ $$=\sum_{i=1}^n\sum_{j=1}^m \left\lfloor\frac ni\right\rfloor \left\lfloor\frac mj\right\rfloor [gcd(i, j) = 1]$$ $$=\sum_{i=1}^{n}\sum_{j=1}^{m}\sum_{d|i}\sum_{d|j}\mu(d)\lfloor\frac{n}{i}\rfloor\lfloor\frac{m}{j}\rfloor$$ $$ =\sum_{d=1}^{min(n,m)}\sum_{i=1}^{\lfloor\frac{n}{d}\rfloor}\sum_{j=1}^{\lfloor\frac{m}{d}\rfloor}\mu(d)\lfloor\frac{n}{i}\rfloor\lfloor\frac{m}{j}\rfloor $$ \]

=\sum_{d=1}{min(n,m)}\mu(d)\sum_{i=1}{\lfloor\frac{n}{d}\rfloor}\lfloor\frac{n}{i}\rfloor\sum_{j=1}^{\lfloor\frac{m}{d}\rfloor}\lfloor\frac{m}{j}\rfloor$$
\(f(x)=\sum_{i=1}^{x}\lfloor\frac{x}{i}\rfloor\)表示1-x的约数个数和

\[\sum_{d=1}^{min(n,m)}\mu(d)f(\lfloor\frac{n}{d}\rfloor)f(\lfloor\frac{m}{d}\rfloor)$$预处理$\mu(n)$前缀和 对于$f$我们可以$O(n\sqrt n)$预处理,查询为$O(\sqrt n)$ ##代码 ```c++ #include<cstdio> #include<cstring> #include<algorithm> const int maxn = 50007; #define int long long inline int read() { int x=0,f=1; char c=getchar(); while(c<'0'||c>'9'){if(c=='0')f=-1;c=getchar();} while(c<='9'&&c>='0')x=x*10+c-'0',c=getchar(); return x*f; } int f[maxn],mu[maxn],num,prime[maxn]; bool vis[maxn]; inline int countf(int x) { int ans=0; for(int i=1,last;i<=x;i=last+1) { last=x/(x/i); ans+=x/i*(last-i+1); } return ans; } void get_mu() { mu[1]=1; for(int i=2;i<maxn;++i) { if(!vis[i])prime[++num]=i,mu[i]=-1; for(int j=1;j<=num&&i*prime[j]<maxn;++j) { vis[prime[j]*i]=1; if(i%prime[j]==0)mu[prime[j]*i]=0; else mu[i*prime[j]]=-mu[i]; } } for(int i=1;i<maxn;++i)mu[i]+=mu[i-1]; } void init() { for(int i=1;i<maxn;++i)f[i]=countf(i); get_mu(); } int solve(int n,int m) { int ans=0; for(int i=1,last;i<=std::min(n,m);i=last+1) { last=std::min(n/(n/i),m/(m/i)); ans+=(mu[last]-mu[i-1])*f[n/i]*f[m/i]; } return ans; } main() { init(); int T=read(); for(;T--;) { printf("%lld\n",solve(read(),read())); } int tmp=0; return 0; } ```\]

posted @ 2018-03-11 13:53  zzzzx  阅读(187)  评论(1编辑  收藏  举报