fzu 2282 Wand

最新推荐文章于 2019-04-12 23:02:33 发布

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最新推荐文章于 2019-04-12 23:02:33 发布

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本文探讨了一个有趣的排列组合问题，即重新排列N个魔法杖，使得至少K位巫师能取回自己的魔法杖，通过动态规划求解该问题，并提供了一种有效的算法实现。

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Problem 2282 Wand

Accept: 8 Submit: 24
Time Limit: 1000 mSec Memory Limit : 262144 KB

Problem Description

N wizards are attending a meeting. Everyone has his own magic wand. N magic wands was put in a line, numbered from 1 to n(Wand_i owned by wizard_i). After the meeting, n wizards will take a wand one by one in the order of 1 to n. A boring wizard decided to reorder the wands. He is wondering how many ways to reorder the wands so that at least k wizards can get his own wand.

For example, n=3. Initially, the wands are w1 w2 w3. After reordering, the wands become w2 w1 w3. So, wizard 1 will take w2, wizard 2 will take w1, wizard 3 will take w3, only wizard 3 get his own wand.

Input

First line contains an integer T (1 ≤ T ≤ 10), represents there are T test cases.

For each test case: Two number n and k.

1<=n <=10000.1<=k<=100. k<=n.

Output

For each test case, output the answer mod 1000000007(10^9 + 7).

Sample Input

21 13 1

Sample Output

n，k ans=∑（cni+dp[n-i]）n>=i>=k;

dp[i]表示i个物品错排的情况数；

错排的状态转移：dp[i]=(i-1)*(dp[i-1]+dp[i-2]);

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
const int MOD=1000000007;
const int N=10007;
long long int qmod(long long int a,long long int b) {
long long int res = 1ll;
while(b) {
if(b&1) res=res*a%MOD;
b>>=1,a=a*a%MOD;
}
return res;
}
long long int dp[N];

long long int fac[N],inv[N];
void init() {
fac[0]=1;
for(long long int i=1; i<N; i++) fac[i]=(fac[i-1]*i)%MOD;
inv[N-1] = qmod(fac[N-1],MOD-2);
for(long long int i=N-2; i>=0; i--) inv[i]=(inv[i+1]*(i+1))%MOD;
}

long long int C(int n,int m) {
return fac[n]*inv[m]%MOD*inv[n-m]%MOD;
}
int main() {
int T;
long long int dp[11111];
int i;
init();
dp[0]=1;
dp[1]=0;
for(i=2;i<=10000;i++)
{
dp[i]=(i-1)*(dp[i-1]+dp[i-2]);
dp[i]=dp[i]%1000000007;
}cin>>T;
while(T--)
{
int a,b;
cin>>a>>b;
long long int ans=0;
for(i=b;i<=a;i++)
{
ans+=dp[a-i]*C(a,i)%1000000007;
ans=ans%1000000007;
}
cout<<ans<<endl;
}
return 0;
}