标签：nbsp max tput scan define hdu class mount answer
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
Hint
There 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}
套路题。
先用dp求出f[i] 为至少有i对满足关系的排列数，然后再二项式反演一下就好啦。
#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int ha=1000000007;
const int maxn=1000;
int jc[maxn+5],ni[maxn+5],n,k,f[maxn+5];
inline int add(int x,int y){ x+=y; return x>=ha?x-ha:x;}
inline int ksm(int x,int y){ int an=1; for(;y;y>>=1,x=x*(ll)x%ha) if(y&1) an=an*(ll)x%ha; return an;}
inline int getC(int x,int y){ return x<y?0:jc[x]*(ll)ni[y]%ha*(ll)ni[x-y]%ha;}
inline void init(){
jc[0]=1;
for(int i=1;i<=maxn;i++) jc[i]=jc[i-1]*(ll)i%ha;
ni[maxn]=ksm(jc[maxn],ha-2);
for(int i=maxn;i;i--) ni[i-1]=ni[i]*(ll)i%ha;
}
inline void solve(){
memset(f,0,sizeof(f));
f[0]=1;
for(int i=n;i;i--)
for(int j=n-i;j>=0;j--) f[j+1]=add(f[j+1],f[j]*(ll)(n-i-j)%ha);
for(int i=0;i<=n;i++) f[i]=f[i]*(ll)jc[n-i]%ha;
int ans=0;
for(int i=k;i<=n;i++)
if((i-k)&1) ans=add(ans,ha-getC(i,k)*(ll)f[i]%ha);
else ans=add(ans,getC(i,k)*(ll)f[i]%ha);
printf("%d\n",ans);
}
int main(){
init();
while(scanf("%d%d",&n,&k)==2) solve();
return 0;
}
HDU - 3664 Permutation Counting
标签：nbsp max tput scan define hdu class mount answer
原文地址：https://www.cnblogs.com/JYYHH/p/8893737.html
