二项式反演的三道练习题

已经没有什么好害怕的了

板子题，所用到的柿子为

\[f(n)=\sum_{i=n}^m\mathrm{C}_i^ng(i)\Leftrightarrow g(n)=\sum_{i=n}^m(-1)^{i-n}\mathrm{C}_i^nf(i) \]

先考虑需要多少组糖果比药片能量大的，就是\(\frac{n+k}{2}\)，这个随便推推就有了。

考虑dp，设\(f_{i,j}\)表示前\(i\)中糖果比恰好糖果比药片能量大的组数钦定\(j\)组的方案数。

记\(r_i\)表示比第\(i\)个糖果能量小的药片的个数。

如果序列无序的话，很不好维护这个东西，发现答案与顺序无关，直接排序即可。

然后dp就好维护了，转移方程为

\[f_{i,j} = f_{i-1,j-1}\times (r_i-j+1) + f_{i-1,j} \]

但是发现这个是不完整的，因为除了钦定的\(j\)个，其它的\(n-j\)个无所谓，要取\(n-j\)的全排列，所以最后要将所有的\(f_{n,i}\)乘上\((n-i)\)的全排列。

然后套二项式反演的柿子求出\(g_k\)即可。

点此查看代码

#include<bits/stdc++.h>
#include<bits/extc++.h>
// using namespace __gnu_pbds;
// using namespace __gnu_cxx;
using namespace std;
#define infile(x) freopen(#x".in","r",stdin)
#define outfile(x) freopen(#x".out","w",stdout)
#define errfile(x) freopen(#x".err","w",stderr)
#define ansfile(x) freopen(#x".ans","w",stdout)
#define rep(i,s,t,p) for(int i = s;i <= t; i += p)
#define drep(i,s,t,p) for(int i = s;i >= t; i -= p)
#ifdef LOCAL
    FILE *InFile = infile(in),*OutFile = outfile(out);
    // FILE *ErrFile = errfile(err);
#else
    FILE *Infile = stdin,*OutFile = stdout;
    //FILE *ErrFile = stderr;
#endif
using ll=long long;using ull=unsigned long long;
using db = double;using ldb = long double;
const int N = 2010,mod = 1e9 + 9;
int n,k,a[N],b[N],f[2][N],fac[N],inv[N],r[N],ans;
//f_{i,j}表示前i个 糖果中比药片能量多的组数 至少为j个的方案数。
inline int power(int a,int b,int mod){
    int res = 1;
    for(;b;b >>= 1,a = 1ll*a*a%mod)
        if(b&1) res = 1ll*res*a%mod;
    return res;
}
inline int Inv(int a){return power(a,mod-2,mod);}
inline int C(int n,int m){return 1ll*fac[n]*inv[m]%mod*inv[n-m]%mod;}
inline void solve(){
    cin>>n>>k;
    if((n-k)%2) return cout<<"0\n",void();
    k = (n+k)>>1;
    rep(i,1,n,1) cin>>a[i];rep(i,1,n,1) cin>>b[i];
    sort(a+1,a+1+n);sort(b+1,b+1+n);
    int now = 0;
    rep(i,1,n,1){
        while(now < n && b[now + 1] < a[i]) ++now;
        r[i] = now;
    }

    fac[0] = 1;
    rep(i,1,n,1) fac[i] = 1ll*fac[i-1]*i%mod;
    inv[n] = Inv(fac[n]);
    drep(i,n-1,0,1) inv[i] = 1ll*inv[i+1]*(i+1)%mod;

    f[0][0] = 1;
    rep(i,1,n,1) rep(j,0,i,1)
        f[i&1][j] = (f[(i-1)&1][j] + (j?1ll*(r[i]-j+1)*f[(i-1)&1][j-1]%mod:0ll))%mod;
    int *dp = f[n&1];
    rep(i,0,n,1) dp[i] = 1ll*dp[i]*fac[n-i]%mod;
    int i = k;
    rep(j,i,n,1){
        int sgn = ((j-i)&1)?-1:1;
        ans = (ans + 1ll*sgn*C(j,i)*dp[j]%mod + mod)%mod;
    }
    cout<<ans<<'\n';
}
signed main(){
    cin.tie(nullptr)->sync_with_stdio(false);
    cout.tie(nullptr)->sync_with_stdio(false);
    solve();
}

集合计数

还是套那个柿子。

\[f(n)=\sum_{i=n}^m\mathrm{C}_i^ng(i)\Leftrightarrow g(n)=\sum_{i=n}^m(-1)^{i-n}\mathrm{C}_i^nf(i) \]

记\(f_i\)表示选出的集合中交集元素个数至少为\(i\)的方案数，有\(f_i=\mathrm{C}_n^i(2^{2^{n-i}})\)

简单解释一下，就是从\(n\)个数中钦定\(i\)个作为交集，然后剩下的数的子集个数为\(2^{n-2}\)，可以选择的方案为\(2^{2^{n-i}}\)个。

然后套柿子容斥就可以了。

点此查看代码

#include<bits/stdc++.h>
#include<bits/extc++.h>
// using namespace __gnu_pbds;
// using namespace __gnu_cxx;
using namespace std;
#define infile(x) freopen(#x".in","r",stdin)
#define outfile(x) freopen(#x".out","w",stdout)
#define errfile(x) freopen(#x".err","w",stderr)
#define ansfile(x) freopen(#x".ans","w",stdout)
#define rep(i,s,t,p) for(int i = s;i <= t; i += p)
#define drep(i,s,t,p) for(int i = s;i >= t; i -= p)
#ifdef LOCAL
    FILE *InFile = infile(in),*OutFile = outfile(out);
    // FILE *ErrFile = errfile(err);
#else
    FILE *Infile = stdin,*OutFile = stdout;
    //FILE *ErrFile = stderr;
#endif
using ll=long long;using ull=unsigned long long;
using db = double;using ldb = long double;
const int N = 1e7 + 10,mod = 1e9 + 7;
int fac[N],inv[N],n,k,ans;
inline int power(int a,int b,int mod){
    int res = 1;
    for(;b;b >>= 1,a = 1ll*a*a%mod)
        if(b&1) res = 1ll*res*a%mod;
    return res;
}
inline int C(int n,int m){return 1ll*fac[n]*inv[m]%mod*inv[n-m]%mod;}
signed main(){
    cin.tie(nullptr)->sync_with_stdio(false);
    cout.tie(nullptr)->sync_with_stdio(false);
    cin>>n>>k;
    fac[0] = 1;
    rep(i,1,n,1) fac[i] = 1ll*fac[i-1]*i%mod;
    inv[n] = power(fac[n],mod-2,mod);
    drep(i,n-1,0,1) inv[i] = 1ll*inv[i+1]*(i+1)%mod;

    rep(i,k,n,1){
        int sgn = ((i-k)&1)?-1:1;
        ans = (ans + 1ll*sgn*C(n,i)*C(i,k)%mod*power(2,power(2,n-i,mod-1),mod)%mod + mod) % mod;
    }
    cout<<ans<<'\n';
}

[NOI Online #2 提高组] 游戏

没错，还是上面那个柿子。~~再粘一遍~~

\[f(n)=\sum_{i=n}^m\mathrm{C}_i^ng(i)\Leftrightarrow g(n)=\sum_{i=n}^m(-1)^{i-n}\mathrm{C}_i^nf(i) \]

设\(f_{x,i}\)表示以\(x\)为根的子树中钦定\(i\)对点会比出胜负，然后树形背包计算，注意最后考虑钦定这个点和子树中的某个节点的情况。

点此查看代码

#include<bits/stdc++.h>
#include<bits/extc++.h>
// using namespace __gnu_pbds;
// using namespace __gnu_cxx;
using namespace std;
#define infile(x) freopen(#x".in","r",stdin)
#define outfile(x) freopen(#x".out","w",stdout)
#define errfile(x) freopen(#x".err","w",stderr)
#define ansfile(x) freopen(#x".ans","w",stdout)
#define rep(i,s,t,p) for(int i = s;i <= t; i += p)
#define drep(i,s,t,p) for(int i = s;i >= t; i -= p)
#ifdef LOCAL
    FILE *InFile = infile(in),*OutFile = outfile(out);
    // FILE *ErrFile = errfile(err);
#else
    FILE *Infile = stdin,*OutFile = stdout;
    //FILE *ErrFile = stderr;
#endif
using ll=long long;using ull=unsigned long long;
using db = double;using ldb = long double;
#define eb emplace_back
const int N = 5010,mod = 998244353;
int n,m,fac[N],inv[N],f[N][N],siz[N][2],g[N];
char s[N];
vector<int> e[N];
inline int power(int a,int b,int mod){
    int res = 1;
    for(;b;b >>= 1,a = 1ll*a*a%mod)
        if(b&1) res = 1ll*res*a%mod;
    return res;
}
inline int Inv(int a){return power(a,mod-2,mod);}
inline int C(int n,int m){return 1ll*fac[n]*inv[m]%mod*inv[n-m]%mod;}
void DP(int x,int fa){
    f[x][0] = 1,siz[x][s[x]-'0']++;
    for(int y:e[x]){
        if(y == fa) continue;
        DP(y,x);
        drep(j,min(m,(siz[x][0]+siz[x][1])/2),0,1){
            drep(k,min((siz[y][0]+siz[y][1])/2,m-j),1,1){
                f[x][j+k] = (f[x][j+k]+1ll*f[x][j]*f[y][k]%mod)%mod;
            }
        }
        siz[x][0] += siz[y][0];
        siz[x][1] += siz[y][1];
    }
    if(s[x] == '1')
        drep(i,min(m-1,siz[x][0]-1),0,1)
            f[x][i+1] = (f[x][i+1] + 1ll*f[x][i]*(siz[x][0]-i)%mod)%mod;
    else 
        drep(i,min(m-1,siz[x][1]-1),0,1)
            f[x][i+1] = (f[x][i+1] + 1ll*f[x][i]*(siz[x][1]-i)%mod)%mod;
}
inline void solve(){
    cin>>n>>(s+1);m = n>>1;
    rep(i,2,n,1){int u,v;cin>>u>>v;e[u].eb(v),e[v].eb(u);}

    fac[0] = 1;
    rep(i,1,m,1) fac[i] = 1ll*fac[i-1]*i%mod;
    inv[m] = Inv(fac[m]);
    drep(i,m-1,0,1) inv[i] = 1ll*inv[i+1]*(i+1)%mod;

    DP(1,0);
    rep(i,0,m,1) g[i] = 1ll*f[1][i]*fac[m-i]%mod;

    rep(i,0,m,1){
        int ans = 0;
        rep(j,i,m,1){
            int sgn = ((j-i)&1)?-1:1;
            ans = (ans + 1ll*sgn*C(j,i)*g[j]%mod + mod)%mod;
        }
        cout<<ans<<'\n';
    }
}
signed main(){
    cin.tie(nullptr)->sync_with_stdio(false);
    cout.tie(nullptr)->sync_with_stdio(false);
    solve();
}

posted @ 2024-10-16 21:40 CuFeO4 阅读(20) 评论(0) 编辑收藏举报

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二项式反演的三道练习题

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