大数相乘-写的比较麻烦

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


void strtoint(char *str, int * cheng, int n)
{
int i;
for(i = 0;i<n;i++)
{
cheng[n-i-1] = str[i] -'0';
}
for(i = 0;i<n;i++)
{
printf("%d ",cheng[i]);
}
printf("\n");
}
int  getresult(int *cheng1,int * cheng2,int len1,int len2,int *sum)
{
int i,j;
int tmp = 0;


for(i = 0;i<len1+len2;i++)
{
sum[i] = 0;
}


for(i = 0 ;i<len2;i++)
{
for(j=0;j<len1;j++)
{
sum[i+j] +=  (cheng2[i] * cheng1[j] %10 + tmp);
printf("sum1[%d] = %d\n",i+j,sum[i+j]);
int tmp1 = sum[i+j];
sum[i+j] = sum[i+j] %10;
printf("sum[%d] = %d\n",i+j,sum[i+j]);
tmp  = tmp1 /10;
printf("tmp = %d\n",tmp);
tmp += cheng2[i] * cheng1[j]/10;
printf("tmp2 = %d\n",tmp);
}
if(tmp != 0) sum[i+j-1] = tmp;


int m;
for(m = 0;m<i+j;m++)
{
printf("%d ",sum[m]);
}
printf("\n");


}
if(tmp == 0) return i+j -2;
else {
return i+j -1   ;
}
}
int main()
{
char str1[1024];
char str2[1024];
char strsum[1024];
int cheng1[1024];
int cheng2[1024];
int sum[1024];
scanf("%s",str1);
scanf("%s",str2);
int len1 = strlen(str1);
int len2 = strlen(str2);
strtoint(str1,cheng1,len1);
strtoint(str2,cheng2,len2);
int num = getresult(cheng1,cheng2,len1,len2,sum);
int i;
printf("num = %d\n",num);
for(i=num;i>=0;i--)
{
printf("%d ",sum[i]);
}
}