Luogu5339 [TJOI2019]唱、跳、rap和篮球 【生成函数，NTT】

当时看到这道题的时候我的脑子可能是这样的：

My left brain has nothing right, and my right brain has nothing left.

总之，看到"没有鸡你太美"这一类就直接想容斥，转化为”选出$i$个鸡你太美“

看到排列问题，直接想指数型生成函数。

设$m=\min(\frac{n}{4},a,b,c,d)$

我们使用万年不变的捆绑法，将鸡你太美当做整体考虑，即在$n-3i$个元素中选$i$个作为鸡你太美，再对其他四种进行全排列。

$$ans=\sum_{i=0}^m(-1)^i(n-4i)!\binom{n-3i}{i}\sum_{i_1\leq a-i}\sum_{i_2\leq b-i}\sum_{i_3\leq c-i}\sum_{i_4\leq d-i}\frac{1}{\prod i_j!}[\sum i_j=n-4i]$$

$$\sum_{i=0}^m(-1)^i\frac{(n-3i)!}{i!}\sum_{i_1\leq a-i}\sum_{i_2\leq b-i}\sum_{i_3\leq c-i}\sum_{i_4\leq d-i}\frac{1}{\prod i_j!}[\sum i_j=n-4i]$$

后面那一长串可以用NTT优化计算。

时间复杂度$O(n^2\log n)$，听说有直接dp的$O(n^2)$做法，但这个生成函数的做法应该是无脑多了。

 #include<bits/stdc++.h>
 #define Rint register int
 using namespace std;
 typedef long long LL;
 const int N =  << , mod = , g = , gi = ;
 inline int kasumi(int a, int b){
     int res = ;
     while(b){
         if(b & ) res = (LL) res * a % mod;
         a = (LL) a * a % mod;
         b >>= ;
     }
     return res;
 }
 int fac[N], inv[N];
 inline void init(int n){
     fac[] = ;
     for(Rint i = ;i <= n;i ++) fac[i] = (LL) i * fac[i - ] % mod;
     inv[n] = kasumi(fac[n], mod - );
     for(Rint i = n;i;i --) inv[i - ] = (LL) inv[i] * i % mod;
 }
 int rev[N];
 inline int calrev(int n){
     int limit = , L = -;
     while(limit <= n){limit <<= ; L ++;}
     for(Rint i = ;i < limit;i ++) rev[i] = (rev[i >> ] >> ) | ((i & ) << L);
     return limit;
 }
 inline void NTT(int *A, int limit, int type){
     for(Rint i = ;i < limit;i ++) if(i < rev[i]) swap(A[i], A[rev[i]]);
     for(Rint mid = ;mid < limit;mid <<= ){
         int Wn = kasumi(type ==  ? g : gi, (mod - ) / (mid << ));
         for(Rint j = ;j < limit;j += mid << ){
             int w = ;
             for(Rint k = ;k < mid;k ++, w = (LL) w * Wn % mod){
                 int x = A[j + k], y = (LL) w * A[j + k + mid] % mod;
                 A[j + k] = (x + y) % mod;
                 A[j + k + mid] = (x - y + mod) % mod;
             }
         }
     }
     if(type == -){
         int inv = kasumi(limit, mod - );
         for(Rint i = ;i < limit;i ++)
             A[i] = (LL) A[i] * inv % mod;
     }
 }
 int n, a, b, c, d, m, ans, A[N], B[N], C[N], D[N];
 int main(){
     scanf("%d%d%d%d%d", &n, &a, &b, &c, &d);
     init(a + b + c + d);
     m = min(n >> , min(min(a, b), min(c, d)));
     for(Rint i = ;i <= m;i ++){
         int limit = calrev(a + b + c + d - (i << ));
         for(Rint j = ;j < limit;j ++){
             A[j] = inv[j] * (j <= a - i);
             B[j] = inv[j] * (j <= b - i);
             C[j] = inv[j] * (j <= c - i);
             D[j] = inv[j] * (j <= d - i);
         }
         NTT(A, limit, ); NTT(B, limit, ); NTT(C, limit, ); NTT(D, limit, );
         for(Rint j = ;j < limit;j ++) A[j] = (LL) A[j] * B[j] % mod * C[j] % mod * D[j] % mod;
         NTT(A, limit, -);
         int tmp = (LL) A[n - (i << )] * fac[n -  * i] % mod * inv[i] % mod;
         if(i & ) ans = (ans - tmp + mod) % mod;
         else ans = (ans + tmp) % mod;
     }
     printf("%d", ans);
 }

Luogu5339