2.1 combination formula
1. C(m,n)=C(m,m-n)
2. C(m,n)=C(m-1,n)+C(m-1,n-1)
derangement D(n) = n!(1 - 1/1! + 1/2! - 1/3! + ... + (-1)^n/n!)
= (n-1)(D(n-2) - D(n-1))
Q(n) = D(n) + D(n-1)
求和公式,k = 1..n
1. sum( k ) = n(n+1)/2
2. sum( 2k-1 ) = n^2
3. sum( k^2 ) = n(n+1)(2n+1)/6
4. sum( (2k-1)^2 ) = n(4n^2-1)/3
5. sum( k^3 ) = (n(n+1)/2)^2
6. sum( (2k-1)^3 ) = n^2(2n^2-1)
7. sum( k^4 ) = n(n+1)(2n+1)(3n^2+3n-1)/30
8. sum( k^5 ) = n^2(n+1)^2(2n^2+2n-1)/12
9. sum( k(k+1) ) = n(n+1)(n+2)/3
10. sum( k(k+1)(k+2) ) = n(n+1)(n+2)(n+3)/4
12. sum( k(k+1)(k+2)(k+3) ) = n(n+1)(n+2)(n+3)(n+4)/5
2.2 generating permutations
// gen_perm lexicographic order to produce P (n-, m) // gen_comb lexicographically generating composition C (n-, m) // gen_perm_swap generate full array transducer adjacent bits P (n-, n-) // generating element 1 used. .n represents // function after generating dummy call, passing a [] and n, a [0] .. a [n-1] is the result of a generated #define MAXN 100 int COUNT; #include <the iostream. H> void dummy ( int * A, int n-) { int I; COUT << << COUNT ++ " : " ; for (I = 0 ; I <N- . 1 ; I ++ ) COUT << A [I] << ' ' ; COUT<<a[n-1]<<endl; } void _gen_perm(int* a,int n,int m,int l,int* temp,int* tag){ int i; if (l==m) dummy(temp,m); else for (i=0;i<n;i++) if (!tag[i]){ temp[l]=a[i],tag[i]=1; _gen_perm(a,n,m,l+1,temp,tag); tag[i]=0; } } void gen_perm(int n,int m){ int a[MAXN],temp[MAXN],tag[MAXN]={0},i; for (i=0;i<n;i++) a[i]=i+1; _gen_perm(a,n,m,0,temp,tag); } void _gen_comb(int* a,int s,int e,int m,int& count,int* temp){ int i; if (!m) dummy(temp,count); else for (i=s;i<=e-m+1;i++){ temp[count++]=a[i]; _gen_comb(a,i+1,e,m-1,count,temp); count--; } } void gen_comb(int n,int m){ int a[MAXN],temp[MAXN],count=0,i; for (i=0;i<n;i++) a[i]=i+1; _gen_comb(a,0,n-1,m,count,temp); } void _gen_perm_swap(int* a,int n,int l,int* pos,int* dir){ int i,p1,p2,t; if (l==n) dummy(a,n); else{ _gen_perm_swap(a,n,l+1,pos,dir); for (i=0;i<l;i++){ p2=(p1=pos[l])+dir[l]; t=a[p1],a[p1]=a[p2],a[p2]=t; pos[a[p1]-1]=p1,pos[a[p2]-1]=p2; _gen_perm_swap(a,n,l+1,pos,dir); } dir[l]=-dir[l]; } } void gen_perm_swap(int n){ int a[MAXN],pos[MAXN],dir[MAXN],i; for (i=0;i<n;i++) a[i]=i+1,pos[i]=i,dir[i]=-1; _gen_perm_swap(a,n,0,pos,dir); }
2.3 generating a gray code
// generates a Gray code the REFLECTED // Gray get the next one yard each call // 000 is the first code 000 ..., 100 ... 000 is the last code void Gray ( int n-, int * code) { int = T 0 , I; for (I = 0 ; I <n-; code = T + [I ++ ]); IF (& T . 1 ) for (n-- ;! code [n-]; N-- ); code [n- - . 1 ] = . 1 the -code [N- . 1 ]; }
2.4 displacement (polya)
// Permutation loop section, polya principle @ Perm [0..n-1] is a permutation 0..n-1 (arrangement) // Returns the substitution cycle minimum, num returns the number of cycles of section #define MAXN 1000 int GCD ( int A, int B) { return B GCD (B, A%? B): A; } int a polyA ( int * Perm, int n-, int & NUM) { int I, J, P, V [ MAXN] = { 0 }, RET = . 1 ; for (NUM = I = 0 ; I <n-; I ++ ) IF (! {V [I]) for (NUM ++, J = 0 , P = I;! V [P = perm [p]]; j ++) V [i] = 1 ; right * = j / gcd (right, j); } Return right; }
2.5 lexicographic full array
// lexicographic whole number and arrangement conversion int perm2num ( int n-, int * P) { int I, J, RET = 0 , K = . 1 ; for (I = N- 2 ; I> = 0 ; K = * N- (i-- )) for (I = J + . 1 ; J <n-; J ++ ) IF (P [J] < P [I]) RET + = K; return RET; } void num2perm ( int n-, int * P, int T) { int I, J; for (I = N- . 1 ; I> =0;i--) p[i]=t%(n-i),t/=n-i; for (i=n-1;i;i--) for (j=i-1;j>=0;j--) if (p[j]<=p[i]) p[i]++; }
2.6 lexicographically composition
// lexicographically combined with the converted number of // COMB is the number of combinations of C (n, m), large numbers replaced when necessary, note processing C (n-, m) = 0 | n-<m int COMB ( int n-, int m) { int RET = . 1 , I; m = m <(nm) m: (nm? ); for (I = nm + . 1 ; I <= n-; RET * = (I ++ )); for ( I = . 1 ; I <= m; RET / = (I ++ )); return m < 0 ? 0 : RET; 58 } int comb2num ( int n-, int m, int * C) { int RET = COMB (n-, m ), i; for (i=0;i<m;i++) ret-=comb(n-c[i],m-i); return ret; } void num2comb(int n,int m,int* c,int t){ int i,j=1,k; for (i=0;i<m;c[i++]=j++) for (;t>(k=comb(n-j,m-i-1));t-=k,j++); }