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hdoj3664Permutation Counting【递推】

2015-11-27 21:04 309 查看

Permutation Counting

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1519    Accepted Submission(s): 769


Problem Description

Given a permutation a1, a2, … aN of {1, 2, …, N}, we define its E-value as the amount of elements where ai > i. For example, the E-value of permutation {1, 3, 2, 4} is 1, while the E-value of {4, 3, 2, 1} is 2. You are requested to find how many permutations
of {1, 2, …, N} whose E-value is exactly k.
 

Input

There are several test cases, and one line for each case, which contains two integers, N and k. (1 <= N <= 1000, 0 <= k <= N). 

 

Output

Output one line for each case. For the answer may be quite huge, you need to output the answer module 1,000,000,007.
 

Sample Input

3 0
3 1

 

Sample Output

1
4

HintThere is only one permutation with E-value 0: {1,2,3}, and there are four permutations with E-value 1: {1,3,2}, {2,1,3}, {3,1,2}, {3,2,1}

 

Source

2010 Asia Regional Harbin
 

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#define MOD 1000000007
using namespace std;
long long f[1010][1010];
void dabiao(){
f[1][1]=1;
f[2][1]=1;f[2][2]=1;
f[3][1]=1;f[3][2]=4;f[3][3]=1;
for(int i=4;i<1010;++i){
f[i][1]=1;
for(int j=2;j<=i;++j){
f[i][j]=(f[i-1][j-1]*(i-j+1)%MOD+(f[i-1][j]*j)%MOD)%MOD;
}
}
}
int main()
{
dabiao();
int n,k;
while(scanf("%d%d",&n,&k)!=EOF){
printf("%lld\n",f
[k+1]);
}
return 0;
}

                                            

                                        

                        
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                                                hdoj3664Permutation
                                                

                        

                    


                

            


            

                
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