【做題】CF177G2. Fibonacci Strings——思維+數列
題意:定義斐波那契字符串為:
- $f_1 = $ "a"
- \(f_2 =\) "b"
- \(f_n = f_{n-1} + f_{n-2}, \, n > 2\)
例如,$f_3 = $ “ba”。
有\(m\)次詢問,第\(i\)次給出一個字符串\(s_i\),問\(s_i\)在\(f_n\)中的出現次數。
\(m \leq 10^4, \, n \leq 10^{18}, \, \sum|s_i| \leq 10^5\)
主要問題在與\(f_p\)與\(f_{p-1}\)拼接時,\(f_p\)的某個後綴與\(f_{p-1}\)的某個前綴可能恰好拼成\(s_i\)
我們設\(f_{p-1}\)長度為\(|s_i|-1\)的前綴為\(a\),長度為\(|s_i|-1\)的後綴為\(b\),\(f_{p}\)長度為\(|s_i|-1\)的前綴為\(a\),長度為\(|s_i|-1\)的後綴為\(c\)。我們觀察發現:
長度為\(|s_i|-1\)的前綴 | 長度為\(|s_i|-1\)的後綴 | 產生額外貢獻的字符串 | |
---|---|---|---|
\(f_{p-1}\) | \(a\) | \(b\) | |
\(f_p\) | \(a\) | \(c\) | \(ca\) |
\(f_{p+1}\) | \(a\) | \(b\) | \(ba\) |
\(f_{p+2}\) | \(a\) | \(c\) | \(ca\) |
\(f_{p+3}\) | \(a\) | \(b\) | \(ba\) |
…… | …… | …… | …… |
\(f_{p+2k}\) | \(a\) | \(c\) | \(ca\) |
\(f_{p+2k+1}\) | \(a\) | \(b\) | \(ba\) |
我們設\(s_i\)在\(ca\)中的出現次數為\(n_c\),在\(ba\)中的出現次數為\(n_b\)
那麽,容易得到
\[O_n = O_{n-1} + O_{n-2} + \begin{cases} n_c, & \text {if $ n\mod 2 = 0$} \\ n_b, & \text{if $n \mod 2 = 1$}\end{cases}\]
考慮拆分貢獻,即設\(A_n\),\(B_n\),\(C_n\)分別表示\(f_n\)中,\(s_i\)在\(f_p\)和\(f_{p+1}\)中的出現次數,在所有\(ba\)中的出現次數,在所有\(ca\)中的出現次數。那麽,我們有
- \(O_n = A_n + B_n \times n_b + C_n \times n_c\)
- \(A_n = A_{n-1} + A_{n-2}\)
- \(B_n = B_{n-1} + B_{n-2} + [n \mod 2 = 1] = B_{n-1} + B_{n-2} + \frac {1 - (-1)^n} {2}\)
- \(C_n = C_{n-1} + C_{n-2} + [n \mod 2 = 0] = C_{n-1} + C_{n-2} + \frac {1 + (-1)^n} {2}\)
- \(B_0 = B_1 = C_0 = C_1 = 0\)
其中,\(A_n\)在我們計算出\(A_0\)和\(A_1\)後,用矩陣快速冪得到。故我們只用考慮\(B_n\)和\(C_n\)這兩個類似的數列。
通過使用OEIS或其他的數列求解方法,我們得到\(B_n = F_{n-1} - \frac {1+(-1)^n}{2}\),以及\(C_n = F_{n} - \frac{1 - (-1)^n}{2}\)。其中,\(F_n\)為第\(n\)個斐波那契數。它們同樣可以用矩陣快速冪求出。
時間復雜度\(O(\sum|s_i| + m \log n )\)。
#include <bits/stdc++.h>
#define int long long
using namespace std;
const int N = 30010, MAX = 200000, MOD = 1000000007;
string fib[N];
int n,m,len,cnt,lef[N],nex[MAX + 10];
char tmp[MAX + 10];
string a,b,c,tp;
struct matrix {
int mat[3][3];
matrix() {
memset(mat,0,sizeof mat);
}
matrix operator * (const matrix& x) const {
matrix ret = matrix();
for (int k = 1 ; k <= 2 ; ++ k)
for (int i = 1 ; i <= 2 ; ++ i)
for (int j = 1 ; j <= 2 ; ++ j)
(ret.mat[i][j] += 1ll * mat[i][k] * x.mat[k][j] % MOD) %= MOD;
return ret;
}
};
matrix bas;
matrix power(matrix a,int b) {
matrix ret = matrix();
ret.mat[1][1] = ret.mat[2][2] = 1;
while (b) {
if (b&1) ret = ret * a;
a = a * a;
b >>= 1;
}
return ret;
}
int getfib(int x) {
matrix ret = power(bas,x);
return ret.mat[1][2];
}
int getnum() {
int ret = 0;
for (int i = 0, j = 0 ; i < (int)tp.length() ; ++ i) {
while (j >= 0 && tmp[j+1] != tp[i])
j = nex[j];
++ j;
if (j == len) ++ ret, j = nex[j];
}
return ret;
}
int solve() {
nex[0] = -1;
for (int i = 2, j = 0 ; i <= len ; ++ i) {
while (j >= 0 && tmp[j+1] != tmp[i])
j = nex[j];
nex[i] = ++j;
}
int p = lower_bound(lef+1,lef+cnt+1,len) - lef;
++ p;
if (n <= p+1) {
tp = fib[n];
return getnum();
}
a = fib[p].substr(0,len-1);
b = fib[p].substr(lef[p] - len+1,len-1);
c = fib[p+1].substr(lef[p+1] - len+1,len-1);
int nb, nc, n0, n1, pos = n - p, ret = 0;
tp = b + a;
nb = getnum();
tp = c + a;
nc = getnum();
tp = fib[p];
n0 = getnum();
tp = fib[p+1];
n1 = getnum();
(ret += 1ll * (getfib(pos) - (pos&1)) * nc % MOD) %= MOD;
(ret += 1ll * (getfib(pos-1) - 1 + (pos&1)) * nb % MOD) %= MOD;
matrix sta = matrix();
sta.mat[1][1] = n1;
sta.mat[1][2] = sta.mat[2][1] = n0;
sta = sta * power(bas,pos);
(ret += sta.mat[1][2]) %= MOD;
ret = (ret % MOD + MOD) % MOD;
return ret;
}
signed main() {
fib[1] = "a";
fib[2] = "b";
for (int i = 3 ; ; ++ i) {
fib[i] = fib[i-1] + fib[i-2];
lef[i] = fib[i].length();
cnt = i;
if (lef[i-1] >= MAX) break;
}
bas.mat[1][1] = bas.mat[1][2] = bas.mat[2][1] = 1;
cin >> n >> m;
for (int i = 1 ; i <= m ; ++ i) {
scanf("%s",tmp+1);
len = strlen(tmp+1);
cout << solve() << endl;
}
return 0;
}
小結:用一種不大簡單的做法做出了這道題。思考時間過長,並且依賴網站來求解數列,這是做此題時體現出的不足之處。
【做題】CF177G2. Fibonacci Strings——思維+數列