完美洗牌算法

//完美洗牌 算法：
#include <iostream>
using namespace std;
#define N 100


void locationreplace(int *buff, int n)
{
int num = 2 * n;
int buff1[N];
int i;
for (i = 1; i <= num; i++)
{
buff1[(i * 2) % (num + 1)] = buff[i];
}
for (i = 1; i <= num; i++)
{
buff[i] = buff1[i];
}
}




//测试程序：
int main()
{
/*int buff1[N];
int sum;
cout << "enter the sum:" << endl;
cin >> sum;
cout << "please input the integer:";
for (int i = 0; i < 2 * sum; i++)
{
cin >> buff1[i];
}*/
int buff[N];
int sum;//sum为所要形成数组的一半
cout << "enter the sum of number to be turned :";
cin >> sum;
for (int i = 1; i <= 2*sum; i++)
{
cin >> buff[i];
}
//int num = sizeof(buff) / sizeof(buff[0]);
//int sum = num / 2;
locationreplace(buff,sum);
for (int i = 1; i <= 2*sum; i++)
{
if (i % 2 != 0)
{
int temp = buff[i];
buff[i] = buff[i + 1];
buff[i + 1] = temp;
}
}
for (int i = 1; i <= 2*sum; i++)
cout << buff[i] << " ";


return 0;
}