出题人很凉心的把算法写成了题目名
首先我们可以发现每一维的贡献是独立的，这可以从 (solve1) 里看出来
然后我们可以考虑转化为 (DP) ，这可以从 (solve2) 里看出来
我们统计每一维能产生的贡献，就是 (a) 个 (0) 面， (b) 个 (1) 面， (c) 个 (2) 面这种形式，能写成一个多项式 (ax^0+bx^1+cx^2),而我们最终显然就是把所有的多项式都乘起来。
暴力一个一个乘就很 naive，分治 (NTT) 解决就好啦。
不透彻的话把每个 (solve) 都看一遍就好啦。
Warning：请不要学习本代码的分治 (NTT) 写法，考场上现想出来的，实现麻烦了不少，建议学一下其他大佬的写法。
#include<iostream>
#include<cstdio>
#define int long long
#define LL long long
using namespace std;
int n;
const int N = 400010, mod = 469762049, G = 3, Ginv = (mod + 1) / 3;
int a[N], b[N], c[N], ans[N];
inline int read()
{
int res = 0; char ch = getchar(); bool XX = false;
for (; !isdigit(ch); ch = getchar())(ch == '-') && (XX = true);
for (; isdigit(ch); ch = getchar())res = (res << 3) + (res << 1) + (ch ^ 48);
return XX ? -res : res;
}
void solve1()
{
int tmp;
for (int i = 1; i <= a[1]; ++i)
{
tmp = 0;
if (i == 1)++tmp; if (i == a[1])++tmp;
++ans[tmp];
}
for (int i = 0; i <= 2 * n; ++i)printf("%lld
", ans[i]);
}
void solve2()
{
int tmp;
for (int i = 1; i <= a[1]; ++i)
for (int j = 1; j <= a[2]; ++j)
{
tmp = 0;
if (i == 1)++tmp; if (i == a[1])++tmp;
if (j == 1)++tmp; if (j == a[2])++tmp;
++ans[tmp];
}
for (int i = 0; i <= 2 * n; ++i)printf("%lld
", ans[i]);
}
void solve3()
{
int tmp;
for (int i = 1; i <= n; ++i)
{
tmp = a[i]; a[i] = b[i] = c[i] = 0;
if (tmp == 1)a[i] = 1, c[i] = 0;
else b[i] = 2, c[i] = tmp - b[i];
}
ans[0] = 1;
for (int i = 1; i <= n; ++i)
{
for (int j = 2 * n; j >= 2; --j)
ans[j] = ((LL)ans[j] * c[i] % mod + (LL)ans[j - 1] * b[i] % mod + (LL)ans[j - 2] * a[i] % mod) % mod;
ans[1] = ((LL)ans[1] * c[i] % mod + (LL)ans[0] * b[i] % mod) % mod;
ans[0] = (LL)ans[0] * c[i] % mod;
}
for (int i = 0; i <= 2 * n; ++i)printf("%lld
", ans[i]);
}
/*下边 solve4*/
int last, top;
int r[N], zhan[30], tmp[500];
LL ksm(LL a, LL b, LL mod)
{
LL res = 1;
for (; b; b >>= 1, a = a * a % mod)
if (b & 1)res = res * a % mod;
return res;
}
void NTT(LL *A, int lim, int opt)
{
if (last != lim)
{
last = lim;
for (int i = 0; i < lim; ++i)
r[i] = (r[i >> 1] >> 1) | (i & 1 ? (lim >> 1) : 0);
}
for (int i = 0; i < lim; ++i)
if (i < r[i])swap(A[i], A[r[i]]);
int len;
LL wn, w, x, y;
for (int mid = 1; mid < lim; mid <<= 1)
{
len = mid << 1;
wn = ksm(opt == 1 ? G : Ginv, (mod - 1) / len, mod);
for (int j = 0; j < lim; j += len)
{
w = 1;
for (int k = j; k < j + mid; ++k, w = w * wn % mod)
{
x = A[k]; y = A[k + mid] * w % mod;
A[k] = (x + y) % mod;
A[k + mid] = (x - y + mod) % mod;
}
}
}
if (opt == 1)return;
int ni = ksm(lim, mod - 2, mod);
for (int i = 0; i < lim; ++i)A[i] = A[i] * ni % mod;
}
void MUL(LL *A, int n, LL *B, int m)
{
if (n + m <= 115)
{
for (int i = 0, to = n + m; i <= to; ++i)tmp[i] = 0;
for (int i = 0; i <= n; ++i)
for (int j = 0; j <= m; ++j)
(tmp[i + j] += A[i] * B[j] % mod) %= mod;
for (int i = 0, to = n + m; i <= to; ++i)A[i] = tmp[i];
for (int i = 0; i <= m; ++i)B[i] = 0;
}
else
{
int lim = 1;
while (lim <= (n + m))lim <<= 1;
NTT(A, lim, 1); NTT(B, lim, 1);
for (int i = 0; i < lim; ++i)A[i] = A[i] * B[i] % mod, B[i] = 0;
NTT(A, lim, -1);
}
}
struct dxs
{
int siz;
LL v[N];
} A[30];
int newdxs()
{
return zhan[top--];
}
void huidxs(int x)
{
A[x].siz = 0;
zhan[++top] = x;
}
int solve(int l, int r)
{
if (l == r)
{
int k = newdxs();
A[k].siz = 2;
A[k].v[0] = c[l]; A[k].v[1] = b[l]; A[k].v[2] = a[l];
return k;
}
int mid = (l + r) >> 1;
int lson = solve(l, mid), rson = solve(mid + 1, r);
MUL(A[lson].v, A[lson].siz, A[rson].v, A[rson].siz);
A[lson].siz = A[lson].siz + A[rson].siz;
huidxs(rson);
return lson;
}
void solve4()
{
int tmp;
for (int i = 1; i <= n; ++i)
{
tmp = a[i]; a[i] = b[i] = c[i] = 0;
if (tmp == 1)a[i] = 1, c[i] = 0;
else b[i] = 2, c[i] = tmp - b[i];
}
for (int i = 1; i <= 25; ++i)zhan[++top] = i;
int k = solve(1, n);
for (int i = 0; i <= 2 * n; ++i)printf("%lld
", A[k].v[i]);
}
/*上边 solve4*/
signed main()
{
freopen("poly.in", "r", stdin);
freopen("poly.out", "w", stdout);
cin >> n;
for (int i = 1; i <= n; ++i)
{
a[i] = read();
}
if (n == 1 && a[1] <= 1000)solve1();
else if (n == 2 && a[1] <= 1000 && a[2] <= 1000)solve2();
else if (n <= 5000)solve3();
else solve4();
fclose(stdin); fclose(stdout);
return 0;
}