c语言实现

好的，下面是使用C语言实现的代码，实现两个长整数的除法和求余操作：

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

#define MAX_LEN 1000

void strrev(char *str) {
    int len = strlen(str);
    for (int i = 0; i < len / 2; i++) {
        char temp = str[i];
        str[i] = str[len - i - 1];
        str[len - i - 1] = temp;
    }
}

void divmod(char *a, char *b, char *q, char *r) {
    int n = strlen(a);
    int m = strlen(b);
    char *A = (char*) malloc(sizeof(char) * (n + 1));
    char *B = (char*) malloc(sizeof(char) * (m + 1));
    strcpy(A, a);
    strcpy(B, b);
    strrev(A);
    strrev(B);
    int *x = (int*) malloc(sizeof(int) * (n + 1));
    int *y = (int*) malloc(sizeof(int) * (m + 1));
    for (int i = 0; i < n; i++) {
        x[i] = A[i] - '0';
    }
    for (int i = 0; i < m; i++) {
        y[i] = B[i] - '0';
    }
    int *q1 = (int*) malloc(sizeof(int) * (n + 1));
    int *r1 = (int*) malloc(sizeof(int) * (n + 1));
    memset(q1, 0, sizeof(int) * (n + 1));
    memset(r1, 0, sizeof(int) * (n + 1));
    for (int i = n - 1; i >= 0; i--) {
        // shift r one bit to left
        for (int j = n - 1; j >= 0; j--) {
            r1[j + 1] = r1[j];
        }
        r1[0] = x[i];
        // binary search to find the quotient
        int l = 0, r = 9;
        while (l < r) {
            int mid = (l + r + 1) / 2;
            int flag = 1;
            for (int j = m - 1; j >= 0; j--) {
                int t = y[j] * mid + r1[j];
                if (t >= 10) {
                    flag = 0;
                    break;
                }
            }
            if (flag) {
                l = mid;
            } else {
                r = mid - 1;
            }
        }
        // subtract y * l from r
        for (int j = m - 1; j >= 0; j--) {
            int t = y[j] * l;
            r1[j] -= t;
            r1[j + 1] -= r1[j] < 0;
            r1[j] += r1[j] < 0 ? 10 : 0;
        }
        q1[i] = l;
    }
    // copy the result to q and r
    int i, j;
    for (i = n - 1; i >= 0; i--) {
        if (q1[i] != 0) break;
    }
    if (i < 0) {
        q[0] = '0';
        q[1] = '\0';
    } else {
        for (j = 0; j <= i; j++) {
            q[j] = q1[i - j] + '0';
        }
        q[j] = '\0';
    }
    for (i = n - 1; i >= 0; i--) {
        if (r1[i] != 0) break;
    }
    if (i < 0) {
        r[0] = '0';
        r[1] = '\0';
    } else {
        for (j = 0; j <= i; j++) {
            r[j] = r1[i - j] + '0';
        }
        r[j] = '\0';
    }
    free(A);
    free(B);
    free(x);
    free(y);
    free(q1);
    free(r1);
}

int main() {
    char a[MAX_LEN];
    char b[MAX_LEN];
    char q[MAX_LEN];
    char r[MAX_LEN];

    printf("Enter the first number: ");
    scanf("%s", a);
    printf("Enter the second number: ");
    scanf("%s", b);

    divmod(a, b, q, r);

    printf("Quotient: %s\n", q);
    printf("Remainder: %s\n", r);

    return 0;
}

注意，这里使用了高精度计算方法，需要动态分配内存来存储数字的每一位。另外，这里的除法操作是整除，即只保留商的整数部分，并且不进行四舍五入。如果需要实现浮点数除法，还需要进行额外的处理。