Any solution to convert given char*(number) to required unsigned char*(hex of number)

Given For e.g,
char input[] = "10011210582553796";

now Hexadecimal of 10011210582553796 is 2391249A8B74C4.

So output unsigned char* should have value,
unsigned char output[8] = {0x00,0x23,0x91,0x24,0x9A,0x8B,0x74,0xC4};

Any solution to convert given input char* to output char* with above mentioned requirement.

code should run on platform for which __int64/signed long long is not supported.

Thanx in advance...

For the format conversion you can use something like sprintf, but for the data size you will have a problem unless you write a routine to convert numbers that large, also you will need to think about whether the length is fixed or variable.

found solution

int number = 0;
char numberchars[] = "10011210582553796";
int i = 0;
int ans = 0;
int carry = 0;

char answerArray[100] = {0};

char remainderArray[100] = {0};
int remindex = 0;
int ansindex = 0;
int remainder = 0;

while( numberchars[i] != '\0' )
{
while( numberchars[i] != '\0' )
{
char currentchar[2] = {0};
currentchar[0]=numberchars[i];

int num = atoi(currentchar);
num += remainder;
remainder = 0;

if ( num < 2 )
{
remainder = num;
if(i>0)
answerArray[ansindex++] = '0';
}
else
{
remainder = num % 2;
int answer = num / 2;

char a[2] = {0};
itoa(answer,a,10);

answerArray[ansindex++] = a[0];
}

i++;
remainder *= 10;
}

char a[2] = {0};
int rval = remainder / 10;
itoa(remainder / 10,a,10);

remainderArray[remindex++] = a[0];
int size = sizeof(answerArray);
memcpy(numberchars,answerArray,sizeof(answerArray) );
size = sizeof(answerArray);
memset(answerArray,0,sizeof(answerArray));
ansindex = 0;
remainder = 0;
i=0;
}

char int64[8] = {0};

for(int k=0;remainderArray[k]!= '\0';k++)
{
int64[7-(k/8)] |= ((remainderArray[k]-'0') << (k%8));
}