莫比乌斯函数+莫比乌斯反演
几个经典的莫比乌斯反演练习题
先来一个莫比乌斯函数板子
1 int N = 10000000; 2 int not_prim[10000050],prim[10000050]; 3 long long mu[10000050],tot; 4 void getmu(long long n){ 5 not_prim[0] = not_prim[1] = 1; 6 mu[1] = 1; 7 for (long long i = 2 ; i<= n; i++){ 8 if (!not_prim[i]){ 9 mu[i] = -1; 10 prim[++tot] = i; 11 } 12 for (long long j = 1; j <= tot && prim[j]*i <= n ; j++){ 13 not_prim[prim[j]*i] = i; 14 if (i % prim[j] == 0){ 15 mu[prim[j]*i] = 0; 16 break; 17 } 18 mu[prim[j]*i] =-mu[i]; 19 } 20 } 21 }
几个例题
BZOJ2154
#include <bits/stdc++.h> const long long mod = 20101009; const double ex = 1e-10; #define inf 0x3f3f3f3f using namespace std; int T; int N = 10000000; int not_prim[10000050],prim[10000050]; long long mu[10000050],tot; long long sum_mu[10000050]; void getmu(long long n){ not_prim[0] = not_prim[1] = 1; mu[1] = 1; for (long long i = 2 ; i<= n; i++){ if (!not_prim[i]){ mu[i] = -1; prim[++tot] = i; } for (long long j = 1; j <= tot && prim[j]*i <= n ; j++){ not_prim[prim[j]*i] = i; if (i % prim[j] == 0){ mu[prim[j]*i] = 0; break; } mu[prim[j]*i] =-mu[i]; } } for (int i = 1 ; i <= n ; i++){ sum_mu[i] = (sum_mu[i-1] + (mu[i] * i * i % mod) + mod ) % mod; } } inline long long sum(long long m,long long n){ return ((m*(m+1)/2LL) % mod) * ((n*(n+1)/2LL) % mod) % mod; } inline long long F(long long m,long long n){ if (n > m) swap(n,m); long long nex; long long ans = 0; for (long long i = 1LL;i<=n; i = nex + 1LL){ nex = min(n/(n/i),m/(m/i)); long long tmp1 = ( (sum_mu[nex] - sum_mu[i-1]) % mod + mod ) % mod; ans = (ans + ( sum(m/i,n/i) * tmp1 % mod ) )% mod; } return ans; } long long solve(long long n,long long m){ long long ans = 0; if (n > m) swap(n,m); long long nex; for (long long i = 1 ; i<=n ; i = nex+1){ nex = min(m/(m/i),n/(n/i)); long long tmp = ((i+nex)*(nex-i+1LL)/2LL) % mod; ans = (ans + (tmp * F(n/i,m/i)) % mod) % mod; } return ans; } int main(){ long long m,n; scanf("%lld%lld",&m,&n); getmu(min(n,m)); printf("%lld\n",solve(m,n)); }
BZOJ2301
#include <bits/stdc++.h> const long long mod = 1e9+7; const double ex = 1e-10; #define inf 0x3f3f3f3f using namespace std; int not_prim[200050],prim[200050],mu[200050],tot,sum_mu[200050]; long long x; void getmu(int n){ not_prim[0] = not_prim[1] = 1; mu[1] = 1; for (int i = 2 ; i<= n; i++){ if (!not_prim[i]){ mu[i] = -1; prim[++tot] = i; } for (int j = 1; j <= tot && prim[j]*i <= n ; j++){ not_prim[prim[j]*i] = i; if (i % prim[j] == 0){ mu[prim[j]*i] = 0; break; } mu[prim[j]*i] =-mu[i]; } } for (int i = 1 ; i <= n ; i++){ sum_mu[i] = sum_mu[i-1] + mu[i]; } } int solve(int n,int m,int x){ int ans = 0; int nex; if (n>m) swap(n,m); for (int i = 1 ; i*x<=n ; i = nex+1){ nex = min(n/(n/(i*x)),m/(m/(i*x)))/x; ans += (n/(i*x)) * (m/(i*x)) * (sum_mu[nex] - sum_mu[i-1]); } return ans; } int main() { int T; cin >> T; getmu(50050); while (T--){ int a,b,c,d,k; scanf("%d%d%d%d%d",&a,&b,&c,&d,&k); int ans = solve(b,d,k) - solve(a-1,d,k) - solve(b,c-1,k) + solve(a-1,c-1,k); printf("%d\n",ans); } }
BZOJ2440
#include <bits/stdc++.h> const long long mod = 1e9+7; const double ex = 1e-10; #define inf 0x3f3f3f3f using namespace std; int not_prim[200050],prim[2000050],mu[2000050],tot; long long x; void getmu(int n){ not_prim[0] = not_prim[1] = 1; mu[1] = 1; for (int i = 2 ; i<= n; i++){ if (!not_prim[i]){ mu[i] = -1; prim[++tot] = i; } for (int j = 1; j <= tot && prim[j]*i <= n ; j++){ not_prim[prim[j]*i] = i; if (i % prim[j] == 0){ mu[prim[j]*i] = 0; break; } mu[prim[j]*i] =-mu[i]; } } } bool check(long long mid){ long long ans = mid; for (long long i = 2 ; i * i <= mid ;i++ ){ ans += mu[i]*(mid/(i*i)); } return ans >= x; } int main() { int T; scanf("%d",&T); getmu(200000); while (T--){ scanf("%I64d",&x); long long r = 1; long long l = 1e10; long long ans = 1; while (r <= l){ long long mid = ( r + l ) >> 1; if (check(mid)){ ans = mid; l = mid - 1; } else r = mid + 1; } printf("%lld\n",ans); } return 0; }
BZOJ2820
#include <bits/stdc++.h> const long long mod = 1e9+7; const double ex = 1e-10; #define inf 0x3f3f3f3f using namespace std; long long not_prim[10000050],prim[10000050],mu[10000050],tot; void getmu(long long n){ not_prim[0] = not_prim[1] = 1LL; mu[1] = 1LL; for (long long i = 2LL ; i<= n; i++){ if (!not_prim[i]){ mu[i] = -1; prim[++tot] = i; } for (long long j = 1; j <= tot && prim[j]*i <= n ; j++){ not_prim[prim[j]*i] = i; if (i % prim[j] == 0LL){ mu[prim[j]*i] = 0LL; break; } mu[prim[j]*i] =-mu[i]; } } } long long c[10000050]; void init(long long n){ for (long long i = 1; i<=tot ;i++){ for (long long j = prim[i] ; j <= n; j+=prim[i]){ c[j] += mu[j/prim[i]]; } } for (long long i = 1 ; i<=n; i++) c[i] += c[i-1]; } long long solve(long long n,long long m){ long long ans = 0; long long nex; if (n>m) swap(n,m); for (long long i = 1; i<=n ; i=nex+1){ nex = min(n/(n/i),m/(m/i)); ans += (n/i)*(m/i)*(c[nex]-c[i-1]); } return ans; } int main(){ int T; scanf("%d",&T); getmu(10000001); init(10000001); while (T--){ long long n,m; scanf("%lld%lld",&n,&m); printf("%lld\n",solve(n,m)); } }
BZOJ3529
#include <bits/stdc++.h> const unsigned int mod = 1LL<<31; const double ex = 1e-10; #define inf 0x3f3f3f3f using namespace std; int T; int N = 100000; int not_prim[100050],prim[100050],mu[100050],tot; void getmu(int n){ not_prim[0] = not_prim[1] = 1LL; mu[1] = 1LL; for (int i = 2LL ; i<= n; i++){ if (!not_prim[i]){ mu[i] = -1; prim[++tot] = i; } for (int j = 1; j <= tot && prim[j]*i <= n ; j++){ not_prim[prim[j]*i] = i; if (i % prim[j] == 0LL){ mu[prim[j]*i] = 0LL; break; } mu[prim[j]*i] =-mu[i]; } } } struct node{ int i; unsigned int fi; }F[100050]; unsigned int c[100050]; int lowbit(int k){ return k&(-k); } void Modify(int num,unsigned int v){ while (num<=N){ c[num]= c[num] + v ; num+=lowbit(num); } } unsigned int Sum(int num){ long long ans = 0; while (num != 0){ ans=ans + c[num]; num-=lowbit(num); } return ans; } void initF(int n){ for (int i = 1; i<=n ; i++){ for (int j = i; j<=n; j+=i){ F[j].fi=F[j].fi + i; } F[i].i = i; } } struct quer{ int n,m; int a; int id; unsigned int ans; }Q[20022]; inline bool cmpa(quer a,quer b){ return a.a < b.a; } inline bool cmpid(quer a,quer b){ return a.id < b.id; } inline bool cmp(node a,node b){ return a.fi < b.fi; } void solve(int n){ sort(Q+1,Q+T+1,cmpa); sort(F+1,F+N+1,cmp); int j = 0; for (int i = 1; i<=T; i++){ while (j+1<=n&&F[j+1].fi<=Q[i].a){ j++; for (int k = F[j].i ; k<=n ; k+=F[j].i){ Modify(k,F[j].fi * mu[k/F[j].i]); } } int n,m; n=Q[i].n; m=Q[i].m; if (n > m ) swap(m,n); int nex; unsigned int ans = 0; for (int k = 1; k<=n; k = nex+1){ nex = min(n/(n/k),m/(m/k)); ans = ans + (n/k) * (m/k) * (Sum(nex) - Sum(k-1)); } Q[i].ans = ans; } sort(Q+1,Q+T+1,cmpid); } int main(){ getmu(100000); initF(100000); scanf("%d",&T); for (int i = 1; i<=T; i++) scanf("%d%d%d",&Q[i].n,&Q[i].m,&Q[i].a),Q[i].id = i; solve(100000); for (int i = 1; i<=T; i++){ printf("%d\n",Q[i].ans%mod); } return 0; }
BZOJ2693
1 #include <bits/stdc++.h> 2 const int mod = 100000009; 3 const double ex = 1e-10; 4 #define inf 0x3f3f3f3f 5 using namespace std; 6 int N = 10000000; 7 int not_prim[10000050],prim[10000050]; 8 int mu[10000050],tot; 9 int pre[10000050]; 10 namespace fastIO{ 11 #define BUF_SIZE 100000 12 #define OUT_SIZE 100000 13 #define ll long long 14 //fread->read 15 bool IOerror=0; 16 inline char nc(){ 17 static char buf[BUF_SIZE],*p1=buf+BUF_SIZE,*pend=buf+BUF_SIZE; 18 if (p1==pend){ 19 p1=buf; pend=buf+fread(buf,1,BUF_SIZE,stdin); 20 if (pend==p1){IOerror=1;return -1;} 21 //{printf("IO error!\n");system("pause");for (;;);exit(0);} 22 } 23 return *p1++; 24 } 25 inline bool blank(char ch){return ch==' '||ch=='\n'||ch=='\r'||ch=='\t';} 26 inline int read(int &x){ 27 bool sign=0; char ch=nc(); x=0; 28 for (;blank(ch);ch=nc()); 29 if (IOerror)return 0; 30 if (ch=='-')sign=1,ch=nc(); 31 for (;ch>='0'&&ch<='9';ch=nc())x=x*10+ch-'0'; 32 if (sign)x=-x; 33 return 1; 34 } 35 inline int read(ll &x){ 36 bool sign=0; char ch=nc(); x=0; 37 for (;blank(ch);ch=nc()); 38 if (IOerror)return 0; 39 if (ch=='-')sign=1,ch=nc(); 40 for (;ch>='0'&&ch<='9';ch=nc())x=x*10+ch-'0'; 41 if (sign)x=-x; 42 return 1; 43 } 44 inline int read(double &x){ 45 bool sign=0; char ch=nc(); x=0; 46 for (;blank(ch);ch=nc()); 47 if (IOerror)return 0; 48 if (ch=='-')sign=1,ch=nc(); 49 for (;ch>='0'&&ch<='9';ch=nc())x=x*10+ch-'0'; 50 if (ch=='.'){ 51 double tmp=1; ch=nc(); 52 for (;ch>='0'&&ch<='9';ch=nc())tmp/=10.0,x+=tmp*(ch-'0'); 53 } 54 if (sign)x=-x; 55 return 1; 56 } 57 inline int read(char *s){ 58 char ch=nc(); 59 for (;blank(ch);ch=nc()); 60 if (IOerror)return 0; 61 for (;!blank(ch)&&!IOerror;ch=nc())*s++=ch; 62 *s=0; 63 return 1; 64 } 65 inline void read(char &c){ 66 for (c=nc();blank(c);c=nc()); 67 if (IOerror){c=-1;return;} 68 } 69 //fwrite->write 70 struct Ostream_fwrite{ 71 char *buf,*p1,*pend; 72 Ostream_fwrite(){buf=new char[BUF_SIZE];p1=buf;pend=buf+BUF_SIZE;} 73 void out(char ch){ 74 if (p1==pend){ 75 fwrite(buf,1,BUF_SIZE,stdout);p1=buf; 76 } 77 *p1++=ch; 78 } 79 void print(int x){ 80 static char s[15],*s1;s1=s; 81 if (!x)*s1++='0';if (x<0)out('-'),x=-x; 82 while(x)*s1++=x%10+'0',x/=10; 83 while(s1--!=s)out(*s1); 84 } 85 void println(int x){ 86 static char s[15],*s1;s1=s; 87 if (!x)*s1++='0';if (x<0)out('-'),x=-x; 88 while(x)*s1++=x%10+'0',x/=10; 89 while(s1--!=s)out(*s1); out('\n'); 90 } 91 void print(ll x){ 92 static char s[25],*s1;s1=s; 93 if (!x)*s1++='0';if (x<0)out('-'),x=-x; 94 while(x)*s1++=x%10+'0',x/=10; 95 while(s1--!=s)out(*s1); 96 } 97 void println(ll x){ 98 static char s[25],*s1;s1=s; 99 if (!x)*s1++='0';if (x<0)out('-'),x=-x; 100 while(x)*s1++=x%10+'0',x/=10; 101 while(s1--!=s)out(*s1); out('\n'); 102 } 103 void print(double x,int y){ 104 static ll mul[]={1,10,100,1000,10000,100000,1000000,10000000,100000000, 105 1000000000,10000000000LL,100000000000LL,1000000000000LL,10000000000000LL, 106 100000000000000LL,1000000000000000LL,10000000000000000LL,100000000000000000LL}; 107 if (x<-1e-12)out('-'),x=-x;x*=mul[y]; 108 ll x1=(ll)floor(x); if (x-floor(x)>=0.5)++x1; 109 ll x2=x1/mul[y],x3=x1-x2*mul[y]; print(x2); 110 if (y>0){out('.'); for (size_t i=1;i<y&&x3*mul[i]<mul[y];out('0'),++i); print(x3);} 111 } 112 void println(double x,int y){print(x,y);out('\n');} 113 void print(char *s){while (*s)out(*s++);} 114 void println(char *s){while (*s)out(*s++);out('\n');} 115 void flush(){if (p1!=buf){fwrite(buf,1,p1-buf,stdout);p1=buf;}} 116 ~Ostream_fwrite(){flush();} 117 }Ostream; 118 inline void print(int x){Ostream.print(x);} 119 inline void println(int x){Ostream.println(x);} 120 inline void print(char x){Ostream.out(x);} 121 inline void println(char x){Ostream.out(x);Ostream.out('\n');} 122 inline void print(ll x){Ostream.print(x);} 123 inline void println(ll x){Ostream.println(x);} 124 inline void print(double x,int y){Ostream.print(x,y);} 125 inline void println(double x,int y){Ostream.println(x,y);} 126 inline void print(char *s){Ostream.print(s);} 127 inline void println(char *s){Ostream.println(s);} 128 inline void println(){Ostream.out('\n');} 129 inline void flush(){Ostream.flush();} 130 }; 131 using namespace fastIO; 132 133 inline void getmu(int n){ 134 not_prim[0] = not_prim[1] = 1; 135 mu[1] = 1; 136 for (int i = 2 ; i<= n; i++){ 137 if (!not_prim[i]){ 138 mu[i] = -1; 139 prim[++tot] = i; 140 } 141 for (int j = 1; j <= tot && prim[j]*i <= n ; j++){ 142 not_prim[prim[j]*i] = i; 143 if (i % prim[j] == 0){ 144 mu[prim[j]*i] = 0; 145 break; 146 } 147 mu[prim[j]*i] =-mu[i]; 148 } 149 } 150 } 151 inline void getpre(int n){ 152 for (int i = 1; i<=n ;i++){ 153 if (mu[i]==0) continue; 154 for (long long j = i ; j<=n ; j+=i){ 155 pre[j] = (pre[j] + mu[i] * i + mod) % mod; 156 } 157 } 158 for (int i = 1; i<=n ;i++){ 159 pre[i] = ( pre[i-1] + (int)((long long)pre[i]*(long long)i % mod ) ) % mod; 160 } 161 } 162 inline long long sum(long long m,long long n){ 163 return ((m*(m+1)/2LL) % mod) * ((n*(n+1)/2LL) % mod) % mod; 164 } 165 inline int solve(int n,int m){ 166 if (n > m) swap(n,m); 167 int nex; 168 int ans = 0; 169 for (long long i = 1 ; i <= n ; i = nex+1){ 170 nex = min(n/(n/i),m/(m/i)); 171 ans = (ans + (int)(( sum(n/i,m/i) * (long long)(pre[nex] - pre[i-1] + mod) ) % mod) ) % mod; 172 } 173 return ans; 174 } 175 int main() 176 { 177 178 int T; 179 scanf("%d",&T); 180 getmu(N); 181 getpre(N); 182 while (T--){ 183 long long n,m; 184 read(n); 185 read(m); 186 print(solve(n,m)); 187 print('\n'); 188 } 189 return 0; 190 }
