分治法求解十进制的两个整数相乘的解C语言

以下是分治法求解十进制的两个整数相乘的解的C语言代码：

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

#define MAX_LEN 1000

void multiply(char x[], char y[], char res[]) {
    int len1 = strlen(x);
    int len2 = strlen(y);

    if (len1 == 0 || len2 == 0) {
        res[0] = '0';
        res[1] = '\0';
        return;
    }

    if (len1 == 1 && len2 == 1) {
        int a = x[0] - '0';
        int b = y[0] - '0';
        int c = a * b;
        if (c < 10) {
            res[0] = c + '0';
            res[1] = '\0';
        } else {
            res[0] = c / 10 + '0';
            res[1] = c % 10 + '0';
            res[2] = '\0';
        }
        return;
    }

    int max_len = len1 > len2 ? len1 : len2;
    if (max_len % 2 != 0) {
        max_len++;
    }

    char a[MAX_LEN], b[MAX_LEN], c[MAX_LEN], d[MAX_LEN];
    char ac[MAX_LEN * 2 + 1], bd[MAX_LEN * 2 + 1], ad_bc[MAX_LEN * 2 + 1];
    memset(a, '0', sizeof(a));
    memset(b, '0', sizeof(b));
    memset(c, '0', sizeof(c));
    memset(d, '0', sizeof(d));
    memset(ac, '0', sizeof(ac));
    memset(bd, '0', sizeof(bd));
    memset(ad_bc, '0', sizeof(ad_bc));

    int i, j;
    for (i = 0; i < len1 / 2; i++) {
        a[i] = x[i];
    }
    for (j = 0; j < len2 / 2; j++) {
        c[j] = y[j];
    }
    multiply(a, c, ac);

    for (i = len1 / 2; i < len1; i++) {
        b[i - len1 / 2] = x[i];
    }
    for (j = len2 / 2; j < len2; j++) {
        d[j - len2 / 2] = y[j];
    }
    multiply(b, d, bd);

    char tmp1[MAX_LEN], tmp2[MAX_LEN];
    memset(tmp1, '0', sizeof(tmp1));
    memset(tmp2, '0', sizeof(tmp2));
    for (i = 0; i < len1 / 2; i++) {
        tmp1[i] = x[i];
    }
    for (j = len2 / 2; j < len2; j++) {
        tmp2[j - len2 / 2] = y[j];
    }
    multiply(tmp1, tmp2, ad_bc);

    for (i = 0; i < max_len * 2; i++) {
        res[i] = '0';
    }
    res[max_len * 2] = '\0';

    for (i = 0; i < strlen(ac); i++) {
        res[i] = ac[i];
    }
    for (i = strlen(ac); i < max_len * 2; i++) {
        res[i] = '0';
    }

    for (i = 0; i < strlen(bd); i++) {
        res[i + max_len * 2 - strlen(bd)] += bd[i] - '0';
    }

    for (i = 0; i < strlen(ad_bc); i++) {
        res[i + max_len] += ad_bc[i] - '0';
    }

    for (i = max_len * 2 - 1; i > 0; i--) {
        if (res[i] > '9') {
            res[i - 1] += (res[i] - '0') / 10;
            res[i] = (res[i] - '0') % 10 + '0';
        }
    }

    while (res[0] == '0' && strlen(res) > 1) {
        for (i = 0; i < strlen(res) - 1; i++) {
            res[i] = res[i + 1];
        }
        res[strlen(res) - 1] = '\0';
    }
}

int main() {
    char x[MAX_LEN], y[MAX_LEN], res[MAX_LEN * 2 + 1];
    printf("请输入两个整数：\n");
    scanf("%s%s", x, y);
    multiply(x, y, res);
    printf("结果为：%s\n", res);
    return 0;
}

其中，multiply函数是分治法求解十进制两个整数相乘的核心部分，采用递归的方式进行计算。主函数中输入两个整数，并调用multiply函数求解它们的乘积。