CodeForces 294C - Shaass and Lights 统计

对于样例3: 1 2 3 4 5 6 7 8 9 10 11

有9个灯要打开...首先处理4前面的...放到9个要打开的任意位置C(9,3)..并且前面几个只有一种操作顺序(从4开始往左)...an=C(9,3)=84

再看5~7..此时还剩下9-3=6个空位置..把这三个放进去..C(6,3)..有5,6,7有四种开的顺序: ->5->6->7; 5<-6<-7<- , ->5->6 7<-,->5 6<-7<-四种..ans*=C(6,3)*4=6720

最后的9~11为C(3,3)只有一种开顺序..所以ans*=C(3,3)=6720...

可知对于被连个亮着的灯夹着的连续灯...若有k个.则有2^(k-1)种开顺序...顶左右两侧的..只能顺序过去...

Program:

#include<iostream>
#include<stack>
#include<queue>
#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<cmath>
#define ll long long
#define oo 1000000007
#define MAXN 1005
using namespace std;    
ll C[MAXN][MAXN],_2jie[MAXN],a[MAXN];
void PreWork()
{
      int i,j;
      for (i=0;i<=1000;i++) C[i][0]=1;
      for (i=1;i<=1000;i++)
         for (j=1;j<=1000;j++)
           C[i][j]=(C[i-1][j]+C[i-1][j-1])%oo;
      _2jie[0]=1;
      for (i=1;i<=1000;i++) _2jie[i]=(_2jie[i-1]*2)%oo;
      return;      
}
int main()
{ 
      int n,m,i;
      ll ans,num;
      PreWork(); 
      while (~scanf("%d%d",&n,&m))
      {
              for (i=1;i<=m;i++) scanf("%I64d",&a[i]);
              sort(a+1,a+1+m);
              num=n-m;
              ans=C[num][a[1]-1],num-=a[1]-1;
              for (i=2;i<=m;i++) 
              {
                    ll x=a[i]-a[i-1]-1;
                    if (x) ans=((ans*C[num][x])%oo*_2jie[x-1])%oo;
                       else ans=(ans*C[num][x])%oo;
                    num-=a[i]-a[i-1]-1;   
              }              
              ans=(ans*C[num][n-a[m]])%oo;
              printf("%I64d\n",ans);
      }
      return 0;
}