1. 程式人生 > >【莫比烏斯函式+除法分塊】BZOJ2301(HAOI2011)[Problem b]題解

【莫比烏斯函式+除法分塊】BZOJ2301(HAOI2011)[Problem b]題解

題目概述

axb,cyd(x,y)=k 的個數。

解題報告

好像很多人說是莫比烏斯反演……但是我感覺只用到了狄利克雷卷積和莫比烏斯函式啊QAQ?

求區間果斷容斥,那麼先寫出答案式子(令 A=Ak,B=Bk ):

i=1Aj=1B[(i,j)=k]i=1Aj=1B[(i,j)=1]i=1Aj=1Be[(i,j)]e=μ1i=1Aj=1Bd|(i,j)μ(d)i=1Aj=1Bd|id|jμ(d)d=1min{A,B}μ(d)AdBd
然後就變成除法分塊了QwQ。

示例程式

#include<cstdio>
#include<algorithm> using namespace std; typedef long long LL; const int maxn=50000; int te,A,B,C,D,K,mu[maxn+5],p[maxn+5];bool pri[maxn+5]; void Make() { pri[1]=true;mu[1]=1; for (int i=2;i<=maxn;i++) { if (!pri[i]) p[++p[0]]=i,mu[i]=-1; for (int j=1,t;j<=p[0]&&(t=i*p[j])<=maxn;j++) { pri[t]=true
;mu[t]=-mu[i]; if (!(i%p[j])) {mu[t]=0;continue;} } } for (int i=2;i<=maxn;i++) mu[i]+=mu[i-1]; } inline LL Solve(int A,int B) { LL ans=0;A/=K;B/=K; for (int l=1,r;l<=A&&l<=B;l=r+1) r=min(A/(A/l),B/(B/l)),ans+=(LL)(mu[r]-mu[l-1])*(A/l)*(B/l); return
ans; } int main() { freopen("program.in","r",stdin); freopen("program.out","w",stdout); for (Make(),scanf("%d",&te);te;te--) { scanf("%d%d%d%d%d",&A,&B,&C,&D,&K);A--;C--; printf("%lld\n",Solve(B,D)-Solve(A,D)-Solve(B,C)+Solve(A,C)); } return 0; }