这是我做的ACM的一道题，有个地方实在想不明白错在哪。。求大神帮看看
先上题：

It was said that when testing the first computer designed by von Neumann, people gave the following problem to both the legendary professor and the new computer: If the 4th digit of 2^n is 7, what is the smallest n? The machine and von Neumann began computing at the same moment, and von Neumann gave the answer first.

Now you are challenged with a similar but more complicated problem: If the K-th digit of M^n is 7, what is the smallest n?

Input

Each case is given in a line with 2 numbers: K and M (< 1,000).

Output

For each test case, please output in a line the smallest n.

You can assume:

The answer always exist.
The answer is no more than 100.
Sample Input

3 2
4 2
4 3
Sample Output

15
21
11

题目很简单，但用到大数乘法，以下是我的代码

#include<stdio.h>

#include<string.h>

#include<stdlib.h>



void reverse(char *s)

{

    int len = strlen(s);

    int i, j;

    char c;

    for (i = 0, j = len - 1; i < j; i++, j--)

    {

        c = *(s + i);

        *(s + i) = *(s + j);

        *(s + j) = c;

    }

}



char *appendTailZero(char *s, int zeros)

{

    int i, len = strlen(s);

    char *r = malloc(len + zeros + 1);

    for (i = 0; i < len; i++)

        *(r + i) = *(s + i);

    for (i = len; i < len + zeros; i++)

        *(r + i) = '0';

    *(r + len + zeros) = '\0';

    return r;

}



char *add(char *s1, char *s2)

{

    int l1 = strlen(s1);

    int l2 = strlen(s2);

    int len = l1 > l2 ? l1 : l2;

    char *r = malloc(len + 2);

    int i, prev = 0, a, b, sum;

    for (i = 0; i < len; i++)

    {

        a = l1 - 1 - i >= 0 ? *(s1 + l1 - 1 - i) - '0' : 0;

        b = l2 - 1 - i >= 0 ? *(s2 + l2 - 1 - i) - '0' : 0;

        sum = a + b + prev;

        *(r + i) = sum > 9 ? sum - 10 + '0' : sum + '0';

        prev = sum > 9 ? 1 : 0;

    }

    if (prev)

    {

        *(r + len) = '1';

        *(r + len + 1) = '\0';

    }

    else

        *(r + len) = '\0';

    reverse(r);

    return r;

}



char *multiplyHelper(char *s1, int digit)

{

    int i, res, prev = 0, len = strlen(s1);

    char *r = malloc(len + 2);

    if (!digit)

    {

        *r = '0';

        *(r + 1) = '\0';

        return r;

    }



    for (i = 0; i < len; i++)

    {

        res = (*(s1 + len - 1 - i) - '0') * digit + prev;

        *(r + i) = res % 10 + '0';

        prev = res / 10;

    }

    if (prev)

    {

        *(r + len) = prev + '0';

        *(r + len + 1) = '\0';

    }

    else

        *(r + len) = '\0';

    reverse(r);

    return r;

}



char *multiply(char *s1, char *s2)

{

    if (strlen(s1) < strlen(s2))

    {

        char *temp;

        temp = s1;

        s1 = s2;

        s2 = temp;

    }

    int l2 = strlen(s2);

    int i;

    char *t, *zt;

    char *sum = malloc(2);

    char *tp;

    sum = "0";

    for (i = 0; i < l2; i++)

    {

        tp = sum;

        t = multiplyHelper(s1, *(s2 + l2 - 1 - i) - '0');

        zt = appendTailZero(t, i);

        sum = add(tp, zt);

        free(t);

        free(zt);

        free(tp);//加入这行代码提交就报Runtime Error, 不加就Accepted,但是内存占用很大

    }

    return sum;

}



int main()

{

    int k;

    char m[5];

    while (scanf("%d %s", &k, m) != EOF)

    {

        char *r = m;

        int count = 1;

        while (strlen(r) < k)

        {

            r = multiply(r, m);

            count++;

        }

        r = r + strlen(r) - k;

        while (*r != '7')

        {

            r = multiply(r, m);

            r = r + strlen(r) - k;

            count++;

        }

        printf("%d\n", count);

    }

    return 0;

}

疑问就出在我注释的那行代码上，由于乘法中间用到了加法，因此每次都会产生加和的临时变量，我想把这些内存free掉，但是一加上这行代码提交就报Runtime Error, 在本地运行没有问题，去掉后提交可以过，但是内存占用很大。想问问这个地方free使用哪里出错了？我free的变量都是前一次的sum结果。。还有最后main函数里面每次做乘法的临时变量内存也可以free，但是加了类似代码也报错。。自己实在想不明白，求大神们帮看看啊，新手分不多见谅