Math Blind Finishing

I. Prime number 1. Sieve prime number: There are two kinds of a linear sieve, a Euler sieve. The general use of Euler sieve on the line, if it is a [l,r] l R large but poor absolute value of a small interval, first with a linear sieve front, and then using Euler sieve sieve behind

　　Euler sieve o (n log log n) : Note that each time I cycle starting from 2 J starting from I

1 BOOLV[n];2 intN;3 voidPrimeintN)4 {5memset (V,0,sizeof(v));6      for(intI=2; i<=n;i++)7     {8         if(V[i])Continue;9cout<<i<<" ";Ten          for(intj=1; j<=n/i;j++) v[i*j]=1; One     } A}

　　linear sieve O (N): J starting from 1

1 voidPrimesintN)2 {3memset (V,0,sizeof(v));4Cnt=0;5      for(intI=2; i<=n;i++)6     {7         if(!v[i]) prime[++cnt]=i,v[i]=i;8          for(intj=1; j<=cnt;j++)9         {Ten             if(prime[j]>v[i]| | prime[j]>n/i) Break; Onev[i*prime[j]]=Prime[j]; A         } -     } -      for(intI=1; i<=cnt;i++) thecout<<prime[i]<<" "; -}

2. Mass factor decomposition: Trial Division. Combined with the 2~ sieve, scan theeach number D of the √n, if D is divisible by n, all factor d is removed from N, and the number of D that is removed is accumulated.

1 voidDivideintN)2 {3Cnt=0;4      for(intI=2; i*i<=n;i++)5     {6         if(n%i==0) 7         {8Prime[++cnt]=i; c[cnt]=0;9              while(n%i==0) n/=i,c[cnt]++;Ten         } One     } A     if(n>1) prime[++cnt]=n,c[cnt]=1; -      for(intI=1; i<=cnt;i++) -cout<<prime[i]<<"^"<<c[i]<<Endl; the}

3.example

1#include <bits/stdc++.h>2 #definell Long Long3 #defineN 100000104 #defineMoD 9982443535 using namespacestd;6 ll Prime[n],cnt,v[n];7 ll L,r,ans,ans2;8 voidPre ()9 {Ten      for(intI=2;i<1000000; i++) One     { A         if(v[i]==0) v[i]=i,prime[++cnt]=i; -          for(intj=1; j<=cnt;j++) -         { the             if(prime[j]>v[i]| | Prime[j]>1000000/i) Break; -v[i*prime[j]]=Prime[j]; -         } -     } + } - intMain () + { A pre (); atscanf"%lld%lld",&l,&R); -memset (V,0,sizeof(v)); -      for(LL i=1; i<=cnt;i++) -     { -ll St=max (1ll, (l1)/prime[i]) *prime[i]+Prime[i]; -          for(LL j=st;j<=r;j+=Prime[i]) in         { -             if(V[j-l])Continue; tov[j-l]=1; +ans++; -Ans2= (Ans2+prime[i])%MoD; the         }    *     } $printf"%lld%lld", ans,ans2);Panax Notoginseng     return 0; -}

Two. Approximate

• Ask for approximate (set)

1. Trial Division: A slight inference: An integer n has an upper bound of approximately several numbers of 2√n2. Multiple methodinference: 1~n The sum of approximately several numbers of each number is approximately n log n

1vector<int>F[n];2 intN;3 voidPuintN)4 {5      for(intI=1; i<=n;i++)6          for(intj=1; j<=n/i;j++)7f[i*J].push_back (i);8      for(intI=1; i<=n;i++){9          for(intj=0; J<f[i].size (); j + +)Tencout<<f[i][j]<<" "; Onecout<<Endl; A     } -}

3.example Inverse prime

See P134

1#include <bits/stdc++.h>2 #defineN 20000000003 #definell Long Long4 using namespacestd;5 ll N;6 ll ans;7 intnum=1;8 inta[ One]={0,2,3,5,7, One, -, -, +, at, in};9 voidDfsintDepintDex,ll Anss,intCNT)Ten { One     if(dep==Ten) A     { -         if((anss>ans&&cnt>num) | | (anss<=ans&&cnt>=num)) -         { theans=Anss; -num=CNT; -         } -         return; +     } -     intt=1; +      for(intI=0; i<=dex;i++) A     { atDFS (dep+1, i,anss*t,cnt* (i+1)); -t*=A[DEP]; -         if(anss*t>n) Break; -     } - } - intMain () in { -scanf"%lld",&n); toDfs1, -,1,1); +printf"%lld\n", ans); -     return 0; the}

• gcd

1. Theorem:

• GCD (A, B) *LCM (A, b) =a*b

• GCD (A, B) =gcd (b,a-b) =GCD (a,a-b) (a>=b)
• GCD (A, a) =gcd (b,a mod b) (b!=0)

　　　 2. Line GCD:

1 int gcd (int A,int  b)2{3      Return B?GCD (b,a%b): A; 4 }

Math Blind Finishing