用C++语言，写出代码实现分治法实现两个n位整数的乘法，并分析该算法的时间复杂度。

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

void add(int *a, int *b, int n) {  // 高精度加法
    int carry = 0;
    for (int i = 0; i < n; i++) {
        carry += a[i] + b[i];
        a[i] = carry % 10;
        carry /= 10;
    }
}

void sub(int *a, int *b, int n) {  // 高精度减法
    int borrow = 0;
    for (int i = 0; i < n; i++) {
        a[i] -= borrow + b[i];
        if (a[i] < 0) {
            a[i] += 10;
            borrow = 1;
        } else {
            borrow = 0;
        }
    }
}

void mul(int *a, int *b, int n, int *c) {  // 高精度乘法
    for (int i = 0; i < n; i++) {
        int carry = 0;
        for (int j = 0; j < n; j++) {
            carry += a[i] * b[j] + c[i + j];
            c[i + j] = carry % 10;
            carry /= 10;
        }
        c[i + n] = carry;
    }
}

void karatsuba(int *a, int *b, int n, int *c) {  // Karatsuba算法
    if (n <= 32) {  // 递归边界
        mul(a, b, n, c);
        return;
    }
    int m = n / 2;
    int *x1 = a, *x2 = a + m, *y1 = b, *y2 = b + m;
    int *z0 = c, *z1 = c + n, *z2 = c + n + m;
    memset(z1, 0, m * 2 * sizeof(int));
    karatsuba(x1, y1, m, z0);         // 计算z0
    karatsuba(x2, y2, m, z2);         // 计算z2
    add(x1, x2, m); add(y1, y2, m);
    karatsuba(x1, y1, m, z1);         // 计算z1
    sub(z1, z0, n); sub(z1, z2, n);
}

int main() {
    char s1[1005], s2[1005];
    scanf("%s%s", s1, s2);
    int n = strlen(s1), m = strlen(s2);
    int len = 1;
    while (len < n + m) len <<= 1;  // 确定多项式的次数
    int *a = (int*)calloc(len, sizeof(int));
    int *b = (int*)calloc(len, sizeof(int));
    int *c = (int*)calloc(len * 2, sizeof(int));
    for (int i = 0; i < n; i++) a[i] = s1[n - 1 - i] - '0';
    for (int i = 0; i < m; i++) b[i] = s2[m - 1 - i] - '0';
    karatsuba(a, b, len, c);
    int k = len * 2 - 1;
    while (k > 0 && c[k] == 0) k--;
    for (int i = k; i >= 0; i--) printf("%d", c[i]);
    printf("\n");
    free(a); free(b); free(c);
    return 0;
}

时间复杂度：设n为两个整数的位数，则Karatsuba算法的时间复杂度为$T(n)=3T(n/2)+O(n)$，其中$O(n)$表示高精度加减乘的时间复杂度。根据主定理，可得$T(n)=O(n^{1.585})$。