标签:nbsp def stdin div 动态 turn for include open
数数题还是要多练啊
code:
#include <cstdio>
#include <cstring>
#include <string>
#include <vector>
#include <algorithm>
#define N 3004
#define ll long long
#define mod 1000000007
#define setIO(s) freopen(s".in","r",stdin)
using namespace std;
int n,D,g[N],f[N][N],s[N][N],fac[N],inv[N];
vector<int>G[N];
int qpow(int x,int y)
{
int tmp=1;
for(;y;y>>=1,x=(ll)x*x%mod)
if(y&1) tmp=(ll)tmp*x%mod;
return tmp;
}
int C(int x,int y) { return (ll)fac[x]*inv[y]%mod*inv[x-y]%mod; }
void dfs(int x)
{
for(int i=1;i<=n;++i) f[x][i]=1;
for(int i=0;i<G[x].size();++i)
{
int y=G[x][i];
dfs(y);
for(int j=1;j<=n;++j) f[x][j]=1ll*f[x][j]*s[y][j]%mod;
}
for(int i=1;i<=n;++i) s[x][i]=(ll)(s[x][i-1]+f[x][i])%mod;
}
int main()
{
// setIO("input");
int i,j;
scanf("%d%d",&n,&D);
for(i=2;i<=n;++i)
{
int ff;
scanf("%d",&ff),G[ff].push_back(i);
}
dfs(1);
fac[0]=inv[0]=1;
for(i=1;i<=n;++i) fac[i]=(ll)i*fac[i-1]%mod, inv[i]=qpow(fac[i],mod-2);
for(i=1;i<=n;++i)
{
g[i]=f[1][i];
for(j=1;j<i;++j)
g[i]=(ll)(g[i]-(ll)C(i-1,i-j)*g[j]%mod+mod)%mod;
}
int ans=0;
int tmp=1;
for(i=1;i<=min(D,n);++i)
{
tmp=(ll)tmp*qpow(i,mod-2)%mod*(D-i+1)%mod;
ans=(ll)(ans+(ll)tmp*g[i]%mod)%mod;
}
printf("%d\n",ans);
return 0;
}
CF995F Cowmpany Cowmpensation 动态规划+容斥原理
标签:nbsp def stdin div 动态 turn for include open
原文地址:https://www.cnblogs.com/guangheli/p/12174561.html