And I'm using the algorithm mentioned here after trying about 10 other ways to do the problem. I'm having trouble though on test case 12 -Consider an ordered set S of strings of N (1 <= N <= 31) bits. Bits, of course, are either 0 or 1.

This set of strings is interesting because it is ordered and contains all possible strings of length N that have L (1 <= L <= N) or fewer bits that are `1'.

Your task is to read a number I (1 <= I <= sizeof(S)) from the input and print the Ith element of the ordered set for N bits with no more than L bits that are `1'.

PROGRAM NAME: kimbits

INPUT FORMAT

A single line with three space separated integers: N, L, and I.

SAMPLE INPUT (file kimbits.in)

5 3 19

OUTPUT FORMAT

A single line containing the integer that represents the Ith element from the order set, as described.

SAMPLE OUTPUT (file kimbits.out)

10011

, I don't know where to start debugging my program ---------------------------

31 26 123456789

--------------------------

Here are the respective outputs:

----- our output ---------

0000111010110111100110100010100

---- your output ---------

0001000001100110000110010110101

--------------------------

Code: Select all

```
#include <stdio.h>
#include <math.h>
FILE *input, *output;
long long numbits, numones, ith, count, total;
long long wrong[2], fact[33], string[33];
long long fac(long long num, long long start)
{
long long i, f=1;
if(fact[num] > 0 && start == 1)
return fact[num];
for(i=start+1; i<= num; i++)
f *= i;
if(start == 1)
fact[num] = f;
return f;
}
int main()
{
long long i, length, numright=0, k, temp=-1, j, numwrong, ornum;
input = fopen("kimbits.in", "r");
output = fopen("kimbits.out", "w");
fscanf(input, "%lld %lld %lld\n", &numbits, &numones, &ith);
ith--;
for(j=1; j<numbits; j++)
{
temp = numbits-j;
numright = 0;
for(i=0; i<=numones && i<=temp; i++) //figure out the number of strings of length temp that fit //the decription
{
if(i > numbits-i)
numright += fac(temp, i)/fac(temp-i, 1);
else
numright += fac(temp, temp-i)/fac(i, 1);
}
if(ith >= numright)
{
ith -= numright;
numones--;
fprintf(output, "1");
}
else
fprintf(output, "0");
}
if(ith >= 1)
{
fprintf(output, "1");
}
else
fprintf(output, "0");
fprintf(output, "\n");
return 0;
}
```