RPG的錯排
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 9816 Accepted Submission(s): 4020
錯排列,組合問題
錯排公式:D(n) = n! [(-1)^2/2! + … + (-1)^(n-1)/(n-1)! + (-1)^n/n!] , D(1) = 0, D(2) = 1
D(n)爲錯排列個數,n爲元素個數
本題需要求0~n/2之間錯排列的總個數
所以, D(1)*c(1, n) + ……+D(k)*c(k, n) +……+ D(n)*c(n, n)
#include <iostream>
typedef long long LL;
using namespace std;
LL f(int i, int n)
{
LL ret = 1;
while (i++ < n)
{
ret *= i;
}
return ret;
}
LL C(int k, int n) //求C(k, n)
{
LL ret = 1;
for (int i = n-k+1; i <= n; i++)
ret *= i;
for (int i = 2; i <= k; i++)
ret /= i;
return ret;
}
int main()
{
int n;
LL sta[15] = {0};
sta[1] = 0;
sta[2] = 1;
for (int i = 3; i < 14; i++)
{
for (int j = 2; j <= i; j++)
{
if (j & 1)
sta[i] -= f(j, i);
else
sta[i] += f(j, i);
}
}
while (cin >> n, n)
{
LL ans = 1;
for (int i = 1; i <= n/2; i++)
ans += C(i, n)*sta[i];
cout << ans << endl;
}
return 0;
}