莫比乌斯反演入门
BZOJ1101
http://www.lydsy.com/JudgeOnline/problem.php?id=1101
求gcd(a,b)=k
a,b中必然包含k的因子
问题等价于a'=a/k,b'=b/k,求有多少对(i,j),其gcd(i,j)=1,1<=i<=a',1<=j<=b'
ans=Σ(i=1,a')Σ(j=1,b')[gcd(i,j)=1]
=Σ(i=1,a')Σ(j=1,b')Σ(d|gcd(i,j))μ(d)
=Σ(i=1,a')Σ(j=1,b')Σ(d|i,d|j)μ(d)
=Σ(d)Σ(i=1,a',d|i)Σ(j=1,b',d|j)μ(d)
=Σ(d)μ(d)*[a'/d]*[b'/d]
这样就有一个线性的做法
复杂度为O(Tn)
70分线性做法
#include<cstdio> typedef long long ll; ll ans; inline int min(int a,int b){ return a<b?a:b; } const int N=50011; bool ip[N]; int pr[N],miu[N]; inline void shai_fa(){ miu[1]=1; for(register int i=2;i<=50000;++i){ if(!ip[i]) miu[pr[++pr[0]]=i]=-1; for(register int j=1;j<=pr[0]&&pr[j]*i<=50000;++j){ ip[pr[j]*i]=1; if(i%pr[j]==0)break; miu[i*pr[j]]=-miu[i]; } } } int T; int a,b,k; int main(){ shai_fa(); scanf("%d",&T); while(T--){ scanf("%d%d%d",&a,&b,&k); a/=k;b/=k; ans=0; for(register int i=1;i<=min(a,b);++i) ans+=1ll*miu[i]*(a/i)*(b/i); printf("%lld\n",ans); } return 0; }
容易知道[a'/d]单调不上升,且最多有2√a'种不同的取值。所以按取值分成√n个段分别处理,一个连续段内的和可以用预处理出的莫比乌斯函数前缀和求出
复杂度O(T√n)
AC根号做法
#include<cstdio> typedef long long ll; ll ans; inline int min(int a,int b){return a<b?a:b;} const int N=50011; bool ip[N]; int pr[N],miu[N],sum[N]; inline void shai_fa(){ sum[1]=miu[1]=1; for(register int i=2;i<=50000;++i){ if(!ip[i]) miu[pr[++pr[0]]=i]=-1; for(register int j=1;j<=pr[0]&&pr[j]*i<=50000;++j){ ip[pr[j]*i]=1; if(i%pr[j]==0)break; miu[i*pr[j]]=-miu[i]; } sum[i]=sum[i-1]+miu[i]; } } int T; int a,b,k,pos; int main(){ shai_fa(); scanf("%d",&T); while(T--){ scanf("%d%d%d",&a,&b,&k); a/=k;b/=k; ans=0; for(register int i=1;i<=min(a,b);i=pos+1){ pos=min(a/(a/i),b/(b/i)); ans+=1ll*(sum[pos]-sum[i-1])*(a/i)*(b/i); } printf("%lld\n",ans); } return 0; }
类似题目BZOJ2301
http://www.lydsy.com/JudgeOnline/problem.php?id=2301
只不过多一个容斥
#include<cstdio> typedef long long ll; ll ans; inline int min(int a,int b){return a<b?a:b;} const int N=50011; bool ip[N]; int pr[N],miu[N],sum[N]; inline void shai_fa(){ sum[1]=miu[1]=1; for(register int i=2;i<=50000;++i){ if(!ip[i]) miu[pr[++pr[0]]=i]=-1; for(register int j=1;j<=pr[0]&&pr[j]*i<=50000;++j){ ip[pr[j]*i]=1; if(i%pr[j]==0)break; miu[i*pr[j]]=-miu[i]; } sum[i]=sum[i-1]+miu[i]; } } int T; int a,b,c,d,k,pos; inline void deal(int n,int m,ll type){ for(register int i=1;i<=min(n,m);i=pos+1){ pos=min(n/(n/i),m/(m/i)); ans+=type*(sum[pos]-sum[i-1])*(n/i)*(m/i); } } int main(){ shai_fa(); scanf("%d",&T); while(T--){ scanf("%d%d%d%d%d",&a,&b,&c,&d,&k); --a;--c; a/=k;b/=k;c/=k;d/=k; ans=0; deal(a,c,1ll);deal(a,d,-1ll);deal(c,b,-1ll);deal(b,d,1ll); printf("%lld\n",ans); } return 0; }