[TJOI2019]唱跳rap和篮球——容斥原理+生成函数

Posted 2022-03-20 dummyummy

先附一组sd图

然后放上原题链接

注意，队伍不同指的是喜好不同，不是人不同

先想到\\(DP\\)，然后你会发现并没有什么优秀的状态设计，然后我们考虑容斥

设\\(lim\\)表示选的癌坤组数的上限，\\(f_i\\)为先选出来\\(i\\)组剩下随便排的方案数，那么答案就是

\\[\\sum\\limits_i=0^lim(-1)^i\\times\\ f_i\\]

于是问题转化为了求\\(f_i\\)。显然\\(f_i\\)可以表示为一个组合数再乘一个东西，具体来说组合数代表在\\(n\\)个同学中选\\(i\\)组癌坤的方案数，后面的那一部分代表满足限制的情况下把\\(n-4i\\)个人随便排的方案数

先考虑把组合数给搞出来，不难发现\\(i\\)组癌坤把\\(n\\)个同学分成了\\(i+1\\)段，这\\(i+1\\)段的长度之和为\\(n-4i\\)，由此我们可以得到以下不定方程

\\[x_1+x_2+...+x_i+1=n-4i\\]

那个组合数就是它的非负整数解个数，这是个经典问题，答案为\\(\\binom(n-4i)+(i+1)-1(i+1)-1=\\binomn-3ii\\)

而后面那一部分的限制条件看起来用生成函数很好搞，同时还是个排列，考虑把\\(4\\)个\\(EGF\\)卷起来，最后后面那一部分的答案即为

\\[[x^n-4i]\\sum\\limits_j=0^a-i\\fracx^jj!\\sum\\limits_j=0^b-i\\fracx^jj!\\sum\\limits_j=0^c-i\\fracx^jj!\\sum\\limits_j=0^d-i\\fracx^jj!\\]

卷积一下就出来了

至此，我们已经得到了一个\\(O(n^2logn)\\)的大常数做法了，但是好像可以在\\(O(n)\\)时间内把后面那一部分求出来（可以参考其他题解）？

放上我的超丑代码：

#include <bits/stdc++.h>

using namespace std;

#define N 1000
#define MOD 998244353

int n, cnta, cntb, cntc, cntd, lim, C[N + 5][N + 5], fac[N + 5], facinv[N + 5];
int ans, a[8 * N + 5], b[8 * N + 5], c[8 * N + 5], d[8 * N + 5];

void add(int &x, int y) 
    x = (x + y) % MOD;
    if (x < 0) x + MOD;


int Add(int x, int y) 
    return ((x + y) % MOD + MOD) % MOD;


int Mul(int x, int y) 
    return (1LL * x * y % MOD + MOD) % MOD;


int fpow(int x, int p) 
    int ret = 1;
    while (p) 
        if (p & 1) ret = Mul(ret, x);
        x = Mul(x, x);
        p >>= 1;
    
    return ret;


void bitReverse(int *s, int len, int bit) 
    int swp[8 * N + 5];
    for (int i = 1; i < len; ++i) 
        swp[i] = (swp[i >> 1] >> 1) | ((i & 1) << (bit - 1));
        if (i < swp[i]) swap(s[i], s[swp[i]]);
    


void NTT(int *s, int len, int bit, int flag) 
    bitReverse(s, len, bit);
    for (int l = 2; l <= len; l <<= 1) 
        int mid = l >> 1, t = fpow(3, (MOD - 1) / l);
        if (flag == -1) t = fpow(t, MOD - 2);
        for (int *p = s; p != s + len; p += l) 
            int w = 1;
            for (int i = 0; i < mid; ++i) 
                int x = 1LL * w * p[i + mid] % MOD;
                p[i + mid] = (p[i] - x) % MOD;
                p[i] = (p[i] + x) % MOD;
                w = 1LL * w * t % MOD;
            
        
    
    if (flag == -1) 
        int invlen = fpow(len, MOD - 2);
        for (int i = 0; i < len; ++i)
            s[i] = 1LL * s[i] * invlen % MOD;
    


int main() 
    scanf("%d%d%d%d%d", &n, &cnta, &cntb, &cntc, &cntd);
    lim = min(n / 4, min(min(cnta, cntb), min(cntc, cntd)));
    for (int i = 0; i <= n; ++i) C[i][0] = 1;
    fac[0] = facinv[0] = 1;
    for (int i = 1; i <= n; ++i) 
        for (int j = 1; j <= i; ++j)
            C[i][j] = Add(C[i - 1][j], C[i - 1][j - 1]);
        fac[i] = Mul(fac[i - 1], i);
        facinv[i] = fpow(fac[i], MOD - 2);
    
    for (int i = 0, l1, l2, l3, l4, len, bit; i <= lim; ++i) 
        int conv;
        for (int j = 0; j <= 8 * N; ++j) a[j] = b[j] = c[j] = d[j] = 0;
        for (int j = 0; j <= cnta - i; ++j) a[j] = facinv[j];
        for (int j = 0; j <= cntb - i; ++j) b[j] = facinv[j];
        for (int j = 0; j <= cntc - i; ++j) c[j] = facinv[j];
        for (int j = 0; j <= cntd - i; ++j) d[j] = facinv[j];
        l1 = cnta - i + 1, l2 = cntb - i + 1, l3 = cntc - i + 1, l4 = cntd - i + 1;
        bit = 0;
        while ((1 << bit) < l1 + l2 - 1) bit++;
        len = 1 << bit;
        NTT(a, len, bit, 1), NTT(b, len, bit, 1);
        for (int j = 0; j < len; ++j) a[j] = Mul(a[j], b[j]);
        NTT(a, len, bit, -1);
        l1 = l1 + l2 - 1;
        bit = 0;
        while ((1 << bit) < l3 + l4 - 1) bit++;
        len = 1 << bit;
        NTT(c, len, bit, 1), NTT(d, len, bit, 1);
        for (int j = 0; j < len; ++j) c[j] = Mul(c[j], d[j]);
        NTT(c, len, bit, -1);
        l2 = l3 + l4 - 1;
        bit = 0;
        while ((1 << bit) < l1 + l2 - 1) bit++;
        len = 1 << bit;
        NTT(a, len, bit, 1), NTT(c, len, bit, 1);
        for (int j = 0; j < len; ++j) a[j] = Mul(a[j], c[j]);
        NTT(a, len, bit, -1);
        conv = a[n - 4 * i];
        if (i & 1) add(ans, -Mul(C[n - 3 * i][i], Mul(fac[n - 4 * i], conv)));
        else add(ans, Mul(C[n - 3 * i][i], Mul(fac[n - 4 * i], conv)));
    
    while (ans < 0) ans += MOD;
    printf("%d\\n", ans);
    return 0;