一些杂项算法

KMP

代码实现

时间复杂度$O(n + m)$

int n, m;
int next[M + 1], f[N + 1];
char s[N + 2], p[M + 2];

void kmp() {
    n = strlen(s + 1), m = strlen(p + 1);
    int j = 0;
    nxt[1] = 0;
    for (int i = 2; i <= m; i++) {
        while (j > 0 && p[j + 1] != p[i])
            j = nxt[i];
        if (p[i + 1] == p[i])
            j++;
        nxt[i] = j;
    }
    j = 0;
    for (int i = 1; i <= n; i++) {
        while ((j == m) || (j > 0 && p[j + 1] != s[i]))
            j = nxt[j];
        if (p[j + 1] == s[i])
            j++;
        f[i] = j;
    }
}
/*
简化写法 两个字符串拼接起来用#分割
    int n = strlen(s+1), m = strlen(p+1);
    p[m + 1] = '#';
    for (int i = 1, j = m + 2; i <= n; i++, j++)
        p[j] = s[i];
    int j = 0;
    nxt[1] = 0;
    for (int i = 2; i <= n + m + 1; i++) {
        while (j && p[i] != p[j + 1])
            j = nxt[j];
        if (p[i] == p[j + 1])
            j++;
        nxt[i] = j;
    }
*/

exkmp

代码实现

时间复杂度$O(n)$

void exkmp() {
	int L = 1, R = 0;
    z[1] = 0;
	for (int i = 2; i <= n; i++) {
        if (l > R) z[i] = 0;
        else {
            int k = i - L + 1;
            z[i] = min(z[k], R - i + 1);
        }
        while (i + z[i] <= n && s[z[i] + 1] == s[i + z[i]])
            z++;
        if (i + z[i] - 1 > R)
            L = i, R = i + z[i] - 1;
    }
}

快速幂

代码实现

时间复杂度$O(logn)$

long long qp(long long a, int n) {
    long long ans = 1;
    for (; n; n >>= 1) {
        if (n & 1) ans *= a, ans %= p;
        a *= a, a %= p;
    }
}

快速乘

代码实现

用于快速计算 $a \times b \pmod p$

当 $0 \le a,b\le10^9,1\le P\le10^9$时,可以直接计算

当 $0 \le a,b\le10^{18},1\le P\le10^9$时,可以$(a\%p \times b\%p)\%p$进行计算

当 $0 \le a,b\le10^{18},1\le P\le10^{18}$时,需要用到快速乘

时间复杂度$O(logn)$

long long qm(long long a, long long b, long long p) {
    long long ans = 0;
    a %= p;//重要
    for (; b; b >>= 1) {
        if (b & 1) ans += a, ans %= p;
        a += a;
        a %= p;
    }
    return ans;
}

矩阵乘法

代码实现

设A是一个n行m列的矩阵,B是一个m行k列的矩阵,矩阵$C=A\times B$

计算方式:$C_{i,j}=\sum_{k=1}^mA_{i,j}B_{k,j}$

A的列数等于B的行数才可以进行矩阵乘法

const int K = 200, P = 1e9 + 7;
int n;
long long a[N + 1][N + 1], f[N + 1];
void aa() {
    long long w[N + 1][N + 1];
    memset(w, 0, sizeof(w));
    for (int i = 1; i <= n; i++) 
        for (int j = 1; j <= n; j++)
            for (int k = 1; k <= n; k++)
                w[i][j] += a[i][k] * a[k][j], w[i][j] %= p;
    memcpy(a, w, sizeof(a));
}

void fa() {
    long long w[N + 1];
    memset(w, 0, sizeof(w));
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++)
            w[i] += f[i] * a[j][i], w[i] %= p;
    memcpy(f, w, sizeof(f));
}

void matrixpow(int k) {
    for (; k; k /= 2) {
        if (k & 1) fa();
        aa();
    }
}