/**//**********************************************
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
*********************************************/
//ploya 定理
//例題：手鐲有n個(gè)位置(位置等距)用來(lái)嵌入寶石，有m種寶石可以用來(lái)嵌入。可以產(chǎn)生多少種不同的手鐲？
//n=4 ,m=3,結(jié)果21
//n=20 ,m=7,結(jié)果1 994 807 299 453 766 (上限)
#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
const int M=20;
int g[M];
int vis[M];
int main()
{
int m,n;
double res=0,tmp;
while(cin>>m>>n)
{
res=0;
int i,j,k,c;
for(i=0;i<m;i++)
{
for(j=0;j<m;j++)
{
g[j]=(j+i)%m;
vis[j]=0;
}
for(j=0,c=0;j<m;j++)
{
if(vis[j]==0)
{
c++;
for(k=j;vis[k]==0;vis[k]=1,k=g[k]);
}
}
res+=pow(n,c);
for(j=1;j*2<m;j++)
{
k=g[j];g[j]=g[m-j];g[m-j]=k;
vis[j]=0;
}
for(j=0;j<m;j++)vis[j]=0;
for(j=0,c=0;j<m;j++)
{
if(vis[j]==0)
{
c++;
for(k=j;vis[k]==0;vis[k]=1,k=g[k]);
}
}
res+=pow(n,c);
}
printf("%0.0lf\n",res/(2*m));
}
return 0;
}
生成排列：n<=8, wei 150ms n==9 wei 1200ms 0(n);
#include<iostream>
#include<ctime>
#include<fstream>
#include <stdio.h>
#include <string.h>
using namespace std;
ifstream fin("g.txt");
ofstream fout("g.out");
int c=0;
void Swap(int &x,int &y)
{
int t;
t=y;
y=x;
x=t;
}
void perm(int list[], int k, int m)
{
if(k==m-1)
{
for(int i=0; i<m; i++) printf("%d",list[i]);
printf("\n");
c++;
}
else
for(int i=k; i<m; i++)
{
Swap(list[k],list[i]);
perm(list,k+1,m);
Swap(list[k],list[i]);
}
}
int main()
{
freopen("g.txt","r",stdin);
freopen("g.out","w",stdout);
int list[100];
memset(list,0,sizeof(list));
int n; scanf("%d",&n);
for(int i=0; i<n;i++)
scanf("%d",&list[i]);
time_t a=clock();
perm(list,0,n);
time_t a1=clock();
printf("%d\n%d\n",a1-a,c);
return 0;
}
高效生成組合算法：
int list[100];
bool b[100];
int n,tot=1;
int C(int n, int m)//計(jì)算組合數(shù)：
{
int result = 1;
if(m > n - m) m = n - m;
for(int i = 1; i <= m; ++i)
{
result = result * (n - m + i) / i;//一定可以整除,^_^
}
return result;
}
void comb()
{
int i,j,k;
while(1)
{
bool c=0;
for(i=1; i<n; i++)
{
if(b[i]==1&&b[i+1]==0)
{
c=1;
break;
}
}
if(c==0) break;
tot++;
b[i+1]=1;
b[i]=0;
for(k=1; k<i; k++)
{
for(j=1; j<i-k; j++)
if(b[j]==0&&b[j+1]==1)
{
bool s;
s=b[j+1];
b[j+1]=b[j];
b[j]=s;
}
}
for(i=1 ;i<=n; i++)
if(b[i]==1)
cout<<list[i];
cout<<endl;
}
}
int main()
{
int m,i;
memset(list,0,sizeof(list));
memset(b,0,sizeof(b));
cin>>n>>m;
for(i=1; i<=n; i++)
cin>>list[i];
for(i=1; i<=m; i++)
{
b[i]=1;
cout<<list[i];
}
cout<<endl;
time_t a=clock();
comb();
time_t a1=clock();
cout<<C(n,m)<<endl;
cout<<tot<<" "<<a1-a<<endl;
return 0;
}
//整數(shù)劃分?jǐn)?shù)的個(gè)數(shù)：
int q(int n, int m)
{
if((n<1)||(m<1)) return 0;
if(n==1||m==1)return 1;
if(n<m) return q(n,n);
if(n==m) return q(n,m-1)+1;
return q(n,m-1)+q(n-m,m);
}
錯(cuò)排的遞推公式：
M(n)=(n-1)[M(n-2)+M(n-1)]
特殊地，M(1)=0,M(2)=1posted on 2010-10-02 14:25
Vontroy 閱讀(2047)
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