[DP 容斥原理] HDU 5519 Kykneion asma

扔下题解就跑
其实我也没全搞懂 先挖个坑

#include<cstdio>#include<cstdlib>#include<algorithm>#include<cstring>using namespace std;typedef long long ll;const int P=1e9+7;const int N=20005;ll fac[N],inv[N];int cnt[1<<5];inline void Pre(int n){  fac[0]=1; for (int i=1;i<=n;i++) fac[i]=fac[i-1]*i%P;  inv[1]=1; for (int i=2;i<=n;i++) inv[i]=(ll)(P-P/i)*inv[P%i]%P;  inv[0]=1; for (int i=1;i<=n;i++) inv[i]=inv[i-1]*inv[i]%P;  for (int i=1;i<(1<<5);i++) cnt[i]=cnt[i>>1]+(i&1);}inline ll C(int n,int m){  return fac[n]*inv[m]%P*inv[n-m]%P;}int n,a[6];int f[N][1<<5];inline void add(int &x,int y){  x+=y; if (x>=P) x-=P;}inline int Solve(){  memset(f,0,sizeof(f));  for (int s=0;s<(1<<5);s++)    f[0][s]=(cnt[s^((1<<5)-1)]&1)?P-1:1;  for (int i=1;i<=n;i++)    for (int s=0;s<(1<<5);s++){      add(f[i][s],(ll)f[i-1][s]*cnt[s]%P);      for (int k=0;k<5;k++)    if ((s>>k&1) && a[k]<i)      add(f[i][s],(ll)f[i-a[k]-1][s^(1<<k)]*C(i-1,a[k])%P);    }  return f[n][(1<<5)-1];}#define read(x) scanf("%d",&(x))int main(){  int T,Case=0;  freopen("t.in","r",stdin);  freopen("t.out","w",stdout);  Pre(20000);  read(T);  while (T--){    read(n); for (int i=0;i<5;i++) read(a[i]);    int ans=Solve();    if (a[0]){      n--; a[0]--;      add(ans,P-Solve());    }    printf("Case #%d: %d\n",++Case,ans);  }  return 0;}