Queuing problem solution composition precision mathematics +

Because it is written does not move, so the chicken dish decadent blog for the future development of the great Austrian letter made little contribution

There are a middle school male students n, m female students and two teachers to queue up to attend a medical examination. They lined up in a straight line, and can not be any adjacent two female students, two teachers can not be adjacent, then a total of how many rows of law do? (Note: any two people are different)

Only one line is separated by spaces and two non-negative integers n and m, its meaning as described above.
For 30% of the data n≤100, m≤100
to 100% of the data n≤2000, m≤2000

Output.txt output file contains only a non-negative integer representing the number of rows of different methods. Note that the answer may be significant.

Sample input

1 1

Sample Output

`12`

题解：
这是一道组合数学，应该说是非常明显。
首先我们需要一个数奥生思路开始要正确。
我们选择固定男生和老师，然后将女生插空，貌似如果固定女生的式子会非常麻烦。我写到一半果断放弃了
固定男生时是A（n,n）;
固定老师时分两种情况:
老师卡在一起，也就是老师一起站在两个男生中间，是A（2，2），这时需要一个女生站在老师中间，总式子变成A(n,n)*(n+1)*A(2,2)*A(n+2,m-1),m-1在上面。
或者老师没有卡在一起,是A（n+1,2）,然后女生插空，总式子变成A(n,n)*A(n+1,2)*A(n+3,m).
合并式子，最终是n!*(2*m+n*(n+3))*(n+1)*(n+2)!/(n-m+3)!。

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #define maxn 100010
 5 using namespace std;
 6 int n,m,a[maxn],b[maxn],c[maxn];
 7 void mult(int a[],int b)
 8 {
 9     int x=0;
10     for(int i=1;i<=a[0];i++)
11     {
12         int tmp=a[i]*b+x;
13         a[i]=tmp%10;
14         x=tmp/10;
15     }
16     while(x)
17     {
18         a[++a[0]]=x%10;
19         x/=10;
20     }
21 }
22 
23 int main()
24 {
25     scanf("%d%d",&n,&m);a[0]=a[1]=1;
26     for(int i=1;i<=n;i++) mult(a,i);
27     mult(a,(2*m+n*(n+3)));
28     mult(a,n+1);
29     for(int i=n-m+4;i<=n+2;i++) mult(a,i);
30     for(int i=a[0];i>0;i--) printf("%d",a[i]);
31     return 0;
32 }

Code

大概就是这样。