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));
}

View Code

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);}
}

View Code

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;
}

View Code

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));}
}

View Code

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;
}

View Code

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 }

卡常数。

转载于:https://www.cnblogs.com/HITLJR/p/7421799.html

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