Solution of Project Euler problems 20 and 26

My recent solution about problem 20 and problem 26 of Project euler

// Problem 20
// Simply implemenent an multiplication of big integer in C++
// and then calculate it

#include 
#include 

#define MAX 100
#define mod 10000
#define baselen 4
#define in(a) scanf("%d",&a)
typedef int type;
/////////////////////////////////////
struct bint{
        type dig[MAX], len;
        bint(){len = 0, dig[0] = 0;}
};


bool setNunmber(bint& a, char number[]){
        type i, j, w, k, p;
        char data[MAX*baselen+1];
        strcpy(data, number);
        w = strlen(data) - 1, a.len = 0;
        for(p=0;p<=w && data[p]=='0';p++);
        while(1){
                i = j = 0, k = 1;
                while(i=p){
                        j = j+ (data[w--] - '0')*k;
                        k *= 10, i++;
                }
                a.dig[a.len++] = j;
                if(w        }
        a.len--;
        return true;
}

void getAnsToString(bint& a, char ans[])
{
        type i;
        i = a.len - 1;
  char *p = ans;
        sprintf(ans, "%d", a.dig[a.len]);
  p += strlen(ans);
        while(i>=0)
  {
     sprintf(p, "%04d",a.dig[i--]);
     p += 4; 
  }
  sprintf(p, "%c", 0);
}



void by(bint a, type b, bint& c){
        type i, carry;
        for( i = carry = 0; i <= a.len || carry; i++){
                if( i <= a.len ) carry += b*a.dig[i];
                c.dig[i] = carry%mod;
                carry /= mod;
        }
        i--;
        while( i && !c.dig[i] )i--;
        c.len = i;
}

int main()
{
 bint product;
 char productInString[MAX*baselen+1];

 setNunmber(product, "1");
 for (int i = 2; i <= 100; ++i) by(product, i, product);
 getAnsToString(product, productInString);

 printf("%s\n", productInString);
 
 int total_digit = 0;
 for (int i = 0; i < strlen(productInString); ++i) total_digit += productInString[i] - '0';
 printf("Sum of the digit = %d\n", total_digit);

 return 0;
}

//Problem 26

#include 
#include 
#include 
using namespace std;
vector recurring_cycle;  //use to record the recurring_cycle

/*  use to record whether an remainder have ever appeard, since when 
 we find an remainder have already exist, we will know we find the 
 recurring cycle. remainder[n] record the first position after radix 
 point we encounter such remainder.
*/
int remainder[1000]; 
int ans_d; // record when will get the max length of recurring cycle
int maxlength = 0; // record the max length of recurring cycle
int main()
{
 for (int d = 2; d < 1000; ++d)
 {
  recurring_cycle.clear();
  memset(remainder, -1, sizeof(remainder));
  int current_remainder = 1;
  int position = 0;
  while (1)
  {
   if (remainder[current_remainder] != -1) 
   {
    /*Uncomment this part to see the recurring cycle
    printf("%d : ", d);
    for (int i = 0; i <= position - 1; ++i) printf("%d", recurring_cycle[i]);
    printf("\n");
    */
    if (position - remainder[current_remainder] > maxlength)
    {
     maxlength = position - remainder[current_remainder];
     ans_d = d;
    }
    break;
   }
   else remainder[current_remainder] = position;
   current_remainder *= 10;
   recurring_cycle.push_back(current_remainder / d);

   if (current_remainder >= d) current_remainder %= d;
   if (current_remainder == 0) 
   {
    /* For test purpose only
    printf("%d : ", d);
    for (int i = 0; i <= position - 1; ++i) printf("%d", recurring_cycle[i]);
    printf("\n");*/
    break;
   }
   ++position;
  }
 }
 printf("Max length of recurring cycle is %d, when d is %d\n", maxlength, ans_d);
 return 0;
}