2020 CCPC-Wannafly Winter Camp Day6 ---A. Convolution

2020 CCPC-Wannafly Winter Camp Day6 ---A. Convolution

这题和2019ICPC南京网络赛的C题很相似。

根据套路，令\(f[i]\)=\(\sum_{k=1}^n [a_k=i]\)
令a中最大值设为n
则\(Ans = \sum_{i=0}^n \sum_{j=0}^n f_if_j2^{ij}\)
= \(2\sum_{i=0}^n \sum_{j=0}^i f_if_j{\sqrt{2}}^{i^2+j^2-{(i-j)}^2}-\sum_{i=0}^n{f_i}^22^{i^2}\)
= \(2\sum_{i=0}^nf_i{{\sqrt{2}}^{i^2}}\sum_{j=0}^if_j{{\sqrt{2}}^{j^2}}{\sqrt{2}}^{{-(i-j)}^2}-\sum_{i=0}^n{f_i}^22^{i^2}\)
令\(b_i=f_i{\sqrt{2}}^{i^2}\) , \(c_i={\sqrt{2}}^{-i^2}\)
则\(b与c卷积成d，其中d中i次项的系数就是\sum_{j=0}^if_j{{\sqrt{2}}^{j^2}}{\sqrt{2}}^{{-(i-j)}^2}\)
所以Ans=\(2\sum_{i=0}^nb_i*d_i-\sum_{i=0}^n{b_i}^2\)
\(d_i\)通过NTT求出即可

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn = 5e5;
const int G = 3;
const ll mol = 998244353;
const ll base = 116195171;
int n,m,L,R[maxn];
ll A[maxn],B[maxn],b[maxn],c[maxn],f[maxn];

ll qpow(ll a,ll b){
    ll ans = 1;
    for (; b; b >>= 1,a = 1ll * a * a % mol)
        if (b & 1) ans = 1ll * ans * a % mol;
    return ans;
}

void NTT(ll* a,int f){
    for (int i = 0; i < n; i++) if (i < R[i]) swap(a[i],a[R[i]]);
    for (int i = 1; i < n; i <<= 1){
        int gn = qpow(G,(mol - 1) / (i << 1));
        for (int j = 0; j < n; j += (i << 1)){
            int g = 1;
            for (int k = 0; k < i; k++,g = 1ll * g * gn % mol){
                int x = a[j + k],y = 1ll * g * a[j + k + i] % mol;
                a[j + k] = (x + y) % mol; a[j + k + i] = (x - y + mol) % mol;
            }
        }
    }
    if (f == 1) return;
    ll nv = qpow(n,mol - 2); reverse(a + 1,a + n);
    for (int i = 0; i < n; i++) a[i] = 1ll * a[i] * nv % mol;
}

ll sub(ll a,ll b) { a -= b; if (a < 0) a += mol; return a; }
ll add(ll a,ll b) { a += b; if (a >= mol) a -= mol; return a; }

int main(){
    scanf("%d" , &n); m = 0;
    for (int i = 1; i <= n; i++) {
    	int x;
    	scanf("%d" , &x);
    	m = max(m , x);
    	f[x]++;
    }
    n = m;
    int tmpn = n;
    for (int i = 0; i <= n; i++) A[i] = b[i] = f[i] * qpow(base , 1ll * i * i) % mol;
    for (int i = 0; i <= n; i++) B[i] = c[i] = qpow(qpow(base , mol - 2) , 1ll * i * i);
    m = n + m; for (n = 1; n <= m; n <<= 1) L++;
    for (int i = 0; i < n; i++) R[i] = (R[i >> 1] >> 1) | ((i & 1) << (L - 1));
    NTT(A,1); NTT(B,1);
    for (int i = 0; i < n; i++) A[i] = 1ll * A[i] * B[i] % mol;
    NTT(A,-1);
    ll ans = 0;
    for (int i = 0; i <= tmpn; i++) ans = add(ans , b[i] * A[i] % mol);
    ans = ans * 2 % mol;
	for (int i = 0; i <= tmpn; i++) ans = sub(ans , b[i] * b[i] % mol);
	printf("%lld\n" , ans);
    return 0;
}