[单位根反演] HDU 7013 String Mod
题目大意
对于所有由前 \(k\) 个小写英文字母组成的长为 \(L\) 的字符串，设 \(A[i][j]\) 表示字符串中 a 的数量模 \(n\) 等于 \(i\)，b 的数量模 \(n\) 等于 \(j\) 这样的字符串个数。输出矩阵 \(A\)。\(2\leq k\leq 26,1\leq L\leq 10^{18},1\leq n\leq 500\)。
题解
设字符串中含有 \(x\) 个 a 和 \(y\) 个 b，对 \(x,y\) 进行枚举，有
\[A[i][j]=\sum_{x=0}^L\sum_{y=0}^{L-x} [n|x-i][n|y-j]\binom{L}{x}\binom{L-x}{y}(k-2)^{L-x-y}\\
\]
考虑换掉艾弗森约定，可以使用单位根反演：
单位根反演
\[\forall k,[n|k]=\frac{1}{n}\sum_{i=0}^{n-1} \omega_n^{ik}
\]
证明：
-
当 \(n|k\) 时，\(\omega_n^{ik}=\omega_n^0=1\)，\(\frac{1}{n}\sum_{i=0}^{n-1}\omega_n^{ik}=1=[n|k]\)
-
当 \(n\nmid k\) 时，
\[\frac{1}{n}\sum_{i=0}^{n-1}\omega_n^{ik}=\frac{1}{n}\cdot\frac{\omega_n^{nk}-1}{\omega_n^k-1}=\frac{1}{n}\cdot\frac{1-1}{\omega_n^k-1}=0=[n|k]
\]
所以有
\[A[i][j]=\sum_{x=0}^L\sum_{y=0}^{L-x} [n|x-i][n|y-j]\binom{L}{x}\binom{L-x}{y}(k-2)^{L-x-y}\\
=\sum_{x=0}^L\sum_{y=0}^{L-x} \left(\frac{1}{n}\sum_{p=0}^{n-1}\omega_n^{p(x-i)}\right)\left(\frac{1}{n}\sum_{q=0}^{n-1}\omega_n^{q(y-j)}\right)\binom{L}{x}\binom{L-x}{y}(k-2)^{L-x-y}\\
=\frac{1}{n^2}\sum_{p=0}^{n-1}\sum_{q=0}^{n-1}\sum_{x=0}^L\sum_{y=0}^{L-x} \omega_n^{p(x-i)}\omega_n^{q(y-j)}\binom{L}{x}\binom{L-x}{y}(k-2)^{L-x-y}\\
=\frac{1}{n^2}\sum_{p=0}^{n-1}\sum_{q=0}^{n-1}\sum_{x=0}^L\sum_{y=0}^{L-x} \omega_n^{px}\omega_n^{qy}\omega_n^{-pi}\omega_n^{-qj}\binom{L}{x}\binom{L-x}{y}(k-2)^L(k-2)^{-x}(k-2)^{-y}\\
=\frac{1}{n^2}\sum_{p=0}^{n-1}\sum_{q=0}^{n-1}\omega_n^{-pi}\omega_n^{-qj}\sum_{x=0}^L\sum_{y=0}^{L-x} \omega_n^{px}\omega_n^{qy}\binom{L}{x}\binom{L-x}{y}(k-2)^{L-x-y}\\
=\frac{1}{n^2}\sum_{p=0}^{n-1}\sum_{q=0}^{n-1}\omega_n^{-pi}\omega_n^{-qj}\sum_{x=0}^L\binom{L}{x}\omega_n^{px}\sum_{y=0}^{L-x} \binom{L-x}{y}(k-2)^{L-x-y}(\omega_n^q)^y\\
=\frac{1}{n^2}\sum_{p=0}^{n-1}\sum_{q=0}^{n-1}\omega_n^{-pi}\omega_n^{-qj}\sum_{x=0}^L\binom{L}{x}(\omega_n^q+k-2)^{L-x}(\omega_n^p)^x\\
=\frac{1}{n^2}\sum_{p=0}^{n-1}\sum_{q=0}^{n-1}\omega_n^{-pi}\omega_n^{-qj}(\omega_n^p+\omega_n^q+k-2)^L
\]
令 \(P(i,p)=\omega_{n}^{-pi},Q(p,q)=(\omega_n^p+\omega_n^q+k-2)^L,R(q,j)=\omega_n^{-qj}\)
则 \(A[i][j]=\frac{1}{n^2}\sum_{p=0}^{n-1}\sum_{q=0}^{n-1}P(i,p)Q(p,q)R(q,j)\)
可以作两次矩阵乘法得到答案。
Code
#include <bits/stdc++.h>
using namespace std;
#define RG register int
#define LL long long
const LL p = 1e9 + 9;
LL qpow(LL b, LL n) {
LL x = 1, Power = b % p;
while (n) {
if (n & 1) x = x * Power % p;
Power = Power * Power % p;
n >>= 1;
}
return x;
}
struct Matrix {
LL a[502][502];
int n;
Matrix operator*(const Matrix& r) {
Matrix res;
res.n = n;
memset(res.a, 0, sizeof(res.a));
for (int k = 0;k < n;++k)
for (int i = 0;i < n;++i)
for (int j = 0;j < n;++j)
res.a[i][j] = (res.a[i][j] + a[i][k] * r.a[k][j] % p) % p;
return res;
}
};
LL x[505], L;
int q[505];
Matrix ans, A, B, C;
int T, k, n;
void solve() {
ans.n = A.n = B.n = C.n = n;
LL g = qpow(13, (p - 1) / n);
x[0] = 1;
for (int i = 1;i < n;++i) x[i] = x[i - 1] * g % p;
for (int i = 0;i < n;++i)
for (int j = 0;j < n;++j) {
A.a[i][j] = qpow(g, i * j);
B.a[i][j] = qpow(x[i] + x[j] + k - 2, L);
}
C = A;
for (int i = 0;i < n;++i)
for (int j = 0;j < i;++j)
swap(C.a[i][j], C.a[j][i]);
ans = A * B;ans = ans * C;
LL m = qpow(n * n, p - 2);
for (int i = 1;i < n;++i) q[i] = n - i;
for (int i = 0;i < n;++i) {
for (int j = 0;j < n;++j) {
printf("%lld", ans.a[q[i]][q[j]] * m % p);
if (j < n - 1) printf(" ");
}
printf("\n");
}
return;
}
int main() {
scanf("%d", &T);
while (T--) {
scanf("%d%lld%d", &k, &L, &n);
solve();
}
return 0;
}
