BZOJ5298 [CQOI2018] 交错序列 | 矩阵乘法和一个trick

题面

$n \le 10^7, m \le 10^8, 0 \le a, b \le 45$

题解

$x^ay^b = (n - y)^ay^b = \sum_{i = 0}^{a}C_a^in^i(-y)^{a - i}y^b = \sum_{i = 0}^{a}(-1)^{a - i}C_a^in^iy^{a+b-i}$

$f[k][i][0/1]$表示长度为$k$、结尾是0/1的序列中“1的个数”（即$y$）的$i$次方之和。

0结尾的序列可以从0/1序列转移过来，而1的出现次数不会变。

$f[k][i][0] = f[k - 1][i][0] + f[k - 1][i][1]$

1结尾的序列只能从0结尾的转移过来，1的出现次数会+1，也就是新的$y' = (y + 1)^i = \sum_{j = 0}^{i}C_i^j y$

$f[k][i][1] = \sum_{j = 0}^{i}C_i^jf[k - 1][j][0]$

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <set>
#define enter putchar('\n')
#define space putchar(' ')
using namespace std;
typedef long long ll;
template <class T>
char c;
bool op = 0;
while(c = getchar(), c < '0' || c > '9')
if(c == '-') op = 1;
x = c - '0';
while(c = getchar(), c >= '0' && c <= '9')
x = x * 10 + c - '0';
if(op == 1) x = -x;
}
template <class T>
void write(T x){
if(x < 0) putchar('-'), x = -x;
if(x >= 10) write(x / 10);
putchar('0' + x % 10);
}

const int N = 185;
int n, a, b, P, sze1, sze2;
ll c[N][N], ans;
struct matrix {
ll g[N][N];
matrix(){
memset(g, 0, sizeof(g));
}
matrix operator * (const matrix &b) const {
matrix c;
for(int i = 0; i < sze2; i++)
for(int j = 0; j < sze2; j++){
for(int k = 0; k < sze2; k++)
c.g[i][j] += g[i][k] * b.g[k][j];
c.g[i][j] %= P;
}
return c;
}
friend matrix qpow(matrix a, int x){
matrix ret;
for(int i = 0; i < sze2; i++)
ret.g[i][i] = 1;
while(x){
if(x & 1) ret = ret * a;
a = a * a;
x >>= 1;
}
return ret;
}
} op;

int main(){
sze1 = a + b + 1, sze2 = 2 * sze1;
c[0][0] = 1;
for(int i = 1; i <= a + b; i++){
c[i][0] = 1;
for(int j = 1; j <= i; j++)
c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % P;
}
for(int i = 0; i < sze1; i++){
op.g[i][i] = op.g[i][sze1 + i] = 1;
for(int j = 0; j <= i; j++)
op.g[sze1 + i][j] = c[i][j];
}
op = qpow(op, n);
ll pw = 1;
for(int i = 0; i <= a; i++){
ans += (((a - i) & 1) ? -1 : 1) * c[a][i] * pw % P * (op.g[a + b - i][0] + op.g[sze1 + a + b - i][0]) % P;
pw = pw * n % P;
}
write((ans % P + P) % P), enter;
return 0;
}

posted @ 2018-04-20 20:04  胡小兔  阅读(488)  评论(0编辑  收藏