康托展开
/*
Name: 康托展开
Copyright: 始发于goal00001111的专栏；允许自由转载，但必须注明作者和出处
Author: goal00001111
Date: 17-11-08 22:50
Description: 康托展开。
把一个整数X展开成如下形式
X=a[n]*n!+a[n-1]*(n-1)!+...+a[2]*2!+a[1]*1!
其中，a为整数，并且0<=a<i,i=1,2,..,n
*/
#include<iostream>
using namespace std;
long long Cantor(int a[], int n);
int main()
{
const int N = 10;
int a[N] = {0};
for (int i=0; i<N; i++)
a[i] = i + 1;
for (int t=0; t<5; t++)
{
for (int i=0; i<N; i++)//随机交换数组元素，得到一个排列
{
int pos = rand() % N;
int temp = a[i];
a[i] = a[pos];
a[pos] = temp;
}
for (int i=0; i<N; i++)
cout << a[i] << ' ';
cout << endl;
cout << Cantor(a, N) << endl << endl;
}
system("pause");
return 0;
}
long long Cantor(int a[], int n)
{
long long s = 1;
long long *p = new long long[n];//用来存储各个全排列的值，p[i]=i!
for (int i=0; i<n-1; i++)
{
s *= i + 1;
p[i] = s;
}
long long sum = 0;
for (int i=0; i<n-1; i++)
{
s = 0;
for (int j=i+1; j<n; j++)
{
if (a[i] > a[j])
s++;
}
sum += s * p[n-2-i];
}
delete []p;
return sum + 1;
}