HDU_4243_Maximum in the Cycle of 1(置换群)

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Maximum in the Cycle of 1

Time Limit:3000MS Memory Limit:0KB 64bit IO Format:%lld & %llu

Submit Status

Description

If P is a permutation of the integers 1,..., n, the maximum in the cycle of 1 is the maximum of the values P(1), P(P(1)), P(P(P(1))), etc. For example, if P is the permutation:

|1 2 3 4 5 6 7 8||3 2 5 4 1 7 8 6|

we have:

P(1) = 3, P(P(1)) = P(3) = 5 and P(P(P(1))) = P(5) = 1

so the maximum in the cycle of 1 is 5.

For this problem, you will write a program which takes as input integers n, (n > 0) and k ( 1 $\le$ k $\le$ n), and returns the number of permutations of the integers 1,..., n, for which the maximum in the cycle of 1 is k.

Input

The first line of inputcontains a single integer P, ( 1 $\le$ P $\le$ 1000), which is the number of data sets that follow. Each data set is a single line that contains the three space separated decimal integer values. The first value is the data set number, N. The second value is the size of the permutation, n where ( 1 $\le$ n $\le$ 20), and the third value is the desired maximum in the cycle of 1, k where ( 1 $\le$ k $\le$ n).

Output

For each data set there is one line of output. It contains the data set number (N) followed by a single space, followed by a double precision floating point whole value which is the number of permutations of the integers 1,..., n, for which the maximum in the xycle of 1 is k.

Sample Input

41 4 12 7 33 10 54 20 7

Sample Output

1 62 1683 864004 11585247657984000

题型：组合数学、离散代数系统

题意：还是不怎么懂的说。。。

分析：貌似用到离散中置换群的概念，还是先去学学离散之后再议，（还是队友比较给力，暂且将代码留下，过会儿慢慢研究）。

代码：

#include<iostream>#include<algorithm>#include<string.h>#include<stack>#include<queue>#include<math.h>#include<cstdio>#define mem(a,b) memset(a,b,sizeof(a))using namespace std;const int maxn=500;const int inf=0x7fffff;double A(double x){double r=1;for(double i=1;i<=x;i++)r*=i;return r;}double C(double m,double n){double r=1;for(double i=n-m+1;i<=n;i++)r*=i;for(double i=m;i>=1;i--)r/=i;return r;}int main(){int t,a;double n,m;scanf("%d",&t);while(t--){scanf("%d%lf%lf",&a,&n,&m);double ans=0;if(m==1) ans+=A(n-1);else for(double i=1;i<m;i++){if(i==1)ans+=A(n-2);else ans+=A(n-1-i)*A(i)*C(i-1,m-2);}printf("%d %0.lf\n",a,ans);}return 0;}