Storing a Big Number in a Variable and Looping - c++

Question

How can i store a big number in a variable and use a for loop?
I have a very big number 75472202764752234070123900087933251 and i need to loop from 0 to this number!
Is it even possible to do this? how much time will it take to end?
EDIT: i am trying to solve a hard problem by brute force. its a combination problem.the bruteforcing cases may reach 470C450.
so i guess i should use a different algorithm...

Answer 1

This might take
0.23 x 10^23 years if C++ processed 100,000 loops per second :|
http://www.wolframalpha.com/input/?i=75472202764752234070123900087933251%2F%28100000*1*3600*24*365%29

Answer 2

It looks that this number fits into 128 bit. So you could use a modern system and a modern compiler that implements such numbers. This would e.g be the case for a 64bit linux system with gcc as a compiler. This has something like __uint128_t that you could use.
Obviously you can't use such a variable as a for-loop variable, others have give you the calculations. But you could use it to store some of your calculations.

Answer 3

Well, you would need an implementation that can handle at least a subset of the initialization, boolean, and arithmetic functions on very large integers. Something like: https://mattmccutchen.net/bigint/.
For something that would give a bit better performance than a general large integer math library, you could use specialized operations specifically to allow use of a large integer as a counter. For an example of this, see dewtell's updated answer to this question.
As for it being possible for you to loop from 0 to that number: well, yes, it is possible to write the code for it with one of the above solutions, but I think the answer is no, you personally will not be able to do it because you will not be alive to see it finish.
[edit: Yes, I would definitely recommend you find a different algorithm. :D]

Answer 4

If you need to loop a certain number of times, and that number is greater than 2^64, just use while(1) because your computer will break before it counts up to 2^64 anyway.

Answer 5

There's no need for a complete bignum package - if all you need is a loop counter, here's a simple byte counter that uses an array of bytes as a counter. It stops when the byte array wraps around to all zeros again. If you wanted to count to some other value than 2^(bytesUsed*CHAR_BITS), you could just compute the two's complement value of the negative of the number of iterations you wanted, and let it count up to 0, keeping in mind that bytes[0] is the low-order byte (or use the positive value and count down instead of up).
#include <stdio.h>
#define MAXBYTES 20
/* Simple byte counter - note it uses argc as # of bytes to use for convenience */
int main(int argc, char **argv) {
unsigned char bytes[MAXBYTES];
const int bytesUsed = argc < MAXBYTES? argc : MAXBYTES;
int i;
unsigned long counter = (unsigned long)-1; /* to give loop something to do */
for (i = 0; i < bytesUsed; i++) bytes[i] = 0; /* Initialize bytes */
do {
for (i = 0; i < bytesUsed && !++bytes[i]; i++) ; /* NULL BODY - this is the byte counter */
counter++;
} while (i < bytesUsed);
printf("With %d bytes used, final counter value = %lu\n", bytesUsed, counter);
}
Run times for the first 4 values (under Cygwin, on a Lenovo T61):
$ time ./bytecounter
With 1 bytes used, final counter value = 255
real 0m0.078s
user 0m0.031s
sys 0m0.046s
$ time ./bytecounter a
With 2 bytes used, final counter value = 65535
real 0m0.063s
user 0m0.031s
sys 0m0.031s
$ time ./bytecounter a a
With 3 bytes used, final counter value = 16777215
real 0m0.125s
user 0m0.015s
sys 0m0.046s
$ time ./bytecounter a a a
With 4 bytes used, final counter value = 4294967295
real 0m6.578s
user 0m0.015s
sys 0m0.047s
At this rate, five bytes should take around half an hour, and six bytes should take the better part of a week. Of course the counter value will be inaccurate for those - it's mostly just there to verify the number of iterations for the smaller byte values and give the loop something to do.
Edit: And here's the time for five bytes, around half an hour as I predicted:
$ time ./bytecounter a a a a
With 5 bytes used, final counter value = 4294967295
real 27m22.184s
user 0m0.015s
sys 0m0.062s

Answer 6

Ok, here's code to take an arbitrary decimal number passed as the first arg and count down from it to zero. I set it up to allow the counter to use different size elements (just change the typedef for COUNTER_BASE), but it turns out that bytes are actually somewhat faster than either short or long on my system.
#include <stdio.h>
#include <limits.h> // defines CHAR_BIT
#include <ctype.h>
#include <vector>
using std::vector;
typedef unsigned char COUNTER_BASE;
typedef vector<COUNTER_BASE> COUNTER;
typedef vector<unsigned char> BYTEVEC;
const unsigned long byteMask = (~0ul) << CHAR_BIT;
const size_t MAXBYTES=20;
void mult10(BYTEVEC &val) {
// Multiply value by 10
unsigned int carry = 0;
int i;
for (i = 0; i < val.size(); i++) {
unsigned long value = val[i]*10ul+carry;
carry = (value & byteMask) >> CHAR_BIT;
val[i] = value & ~byteMask;
}
if (carry > 0) val.push_back(carry);
}
void addDigit(BYTEVEC &val, const char digit) {
// Add digit to the number in BYTEVEC.
unsigned int carry = digit - '0'; // Assumes ASCII char set
int i;
for (i = 0; i < val.size() && carry; i++) {
unsigned long value = static_cast<unsigned long>(val[i])+carry;
carry = (value & byteMask) >> CHAR_BIT;
val[i] = value & ~byteMask;
}
if (carry > 0) val.push_back(carry);
}
BYTEVEC Cstr2Bytevec(const char *str) {
// Turn a C-style string into a BYTEVEC. Only the digits in str apply,
// so that one can use commas, underscores, or other non-digits to separate
// digit groups.
BYTEVEC result;
result.reserve(MAXBYTES);
result[0]=0;
unsigned char *res=&result[0]; // For debugging
while (*str) {
if (isdigit(static_cast<int>(*str))) {
mult10(result);
addDigit(result, *str);
}
str++;
}
return result;
}
void packCounter(COUNTER &ctr, const BYTEVEC &val) {
// Pack the bytes from val into the (possibly larger) datatype of COUNTER
int i;
ctr.erase(ctr.begin(), ctr.end());
COUNTER_BASE value = 0;
for (i = 0; i < val.size(); i++) {
int pos = i%sizeof(COUNTER_BASE); // position of this byte in the value
if (i > 0 && pos == 0) {
ctr.push_back(value);
value = val[i];
} else {
value |= static_cast<COUNTER_BASE>(val[i]) << pos*CHAR_BIT;
}
}
ctr.push_back(value);
}
inline bool decrementAndTest(COUNTER &ctr) {
// decrement value in ctr and return true if old value was not all zeros
int i;
for (i = 0; i < ctr.size() && !(ctr[i]--); i++) ; // EMPTY BODY
return i < ctr.size();
}
inline bool decrementAndTest2(COUNTER_BASE *ctr, const size_t size) {
// decrement value in ctr and return true if old value was not all zeros
int i;
for (i = 0; i < size && !(ctr[i]--); i++) ; // EMPTY BODY
return i < size;
}
/* Vector counter - uses first arg (if supplied) as the count */
int main(int argc, const char *argv[]) {
BYTEVEC limit = Cstr2Bytevec(argc > 1? argv[1] : "0");
COUNTER ctr;
packCounter(ctr, limit);
COUNTER_BASE *ctr_vals = ctr.size() > 0 ? &ctr[0] : NULL;
size_t ctr_size = ctr.size();
unsigned long ul_counter = 0ul; /* to give loop something to do */
while(decrementAndTest2(ctr_vals, ctr_size)) {
ul_counter++;
};
printf("With %d bytes used, final ul_counter value = %lu\n", limit.size(), ul_counter);
return 0;
}
Examples of use:
$ time ./bigcounter 5
With 1 bytes used, final ul_counter value = 5
real 0m0.094s
user 0m0.031s
sys 0m0.047s
$ time ./bigcounter 5,000
With 2 bytes used, final ul_counter value = 5000
real 0m0.062s
user 0m0.015s
sys 0m0.062s
$ time ./bigcounter 5,000,000
With 3 bytes used, final ul_counter value = 5000000
real 0m0.093s
user 0m0.015s
sys 0m0.046s
$ time ./bigcounter 1,000,000,000
With 4 bytes used, final ul_counter value = 1000000000
real 0m2.688s
user 0m0.015s
sys 0m0.015s
$ time ./bigcounter 2,000,000,000
With 4 bytes used, final ul_counter value = 2000000000
real 0m5.125s
user 0m0.015s
sys 0m0.046s
$ time ./bigcounter 3,000,000,000
With 4 bytes used, final ul_counter value = 3000000000
real 0m7.485s
user 0m0.031s
sys 0m0.047s
$ time ./bigcounter 4,000,000,000
With 4 bytes used, final ul_counter value = 4000000000
real 0m9.875s
user 0m0.015s
sys 0m0.046s
$ time ./bigcounter 5,000,000,000
With 5 bytes used, final ul_counter value = 705032704
real 0m12.594s
user 0m0.046s
sys 0m0.015s
$ time ./bigcounter 6,000,000,000
With 5 bytes used, final ul_counter value = 1705032704
real 0m14.813s
user 0m0.015s
sys 0m0.062s
Unwrapping the counter vector into C-style data structures (i.e., using decrementAndTest2 instead of decrementAndTest) sped things up by around 20-25%, but the code is still about twice as slow as my previous C program for similar-sized examples (around 4 billion). This is with MS Visual C++ 6.0 as the compiler in release mode, optimizing for speed, on a 2GHz dual-core system, for both programs. Inlining the decrementAndTest2 function definitely makes a big difference (around 12 sec. vs. 30 for the 5 billion loop), but I'll have to see whether physically inlining the code as I did in the C program can get similar performance.

Answer 7

the variable in main function can Store even 100 factorial
#include <iostream>
#include <cstdio>
#include <vector>
#include <cstring>
#include <string>
#include <map>
#include <functional>
#include <algorithm>
#include <cstdlib>
#include <iomanip>
#include <stack>
#include <queue>
#include <deque>
#include <limits>
#include <cmath>
#include <numeric>
#include <set>
using namespace std;
//template for BIGINIT
// base and base_digits must be consistent
const int base = 10;
const int base_digits = 1;
struct bigint {
vector<int> a;
int sign;
bigint() :
sign(1) {
}
bigint(long long v) {
*this = v;
}
bigint(const string &s) {
read(s);
}
void operator=(const bigint &v) {
sign = v.sign;
a = v.a;
}
void operator=(long long v) {
sign = 1;
if (v < 0)
sign = -1, v = -v;
for (; v > 0; v = v / base)
a.push_back(v % base);
}
bigint operator+(const bigint &v) const {
if (sign == v.sign) {
bigint res = v;
for (int i = 0, carry = 0; i < (int) max(a.size(), v.a.size()) || carry; ++i) {
if (i == (int) res.a.size())
res.a.push_back(0);
res.a[i] += carry + (i < (int) a.size() ? a[i] : 0);
carry = res.a[i] >= base;
if (carry)
res.a[i] -= base;
}
return res;
}
return *this - (-v);
}
bigint operator-(const bigint &v) const {
if (sign == v.sign) {
if (abs() >= v.abs()) {
bigint res = *this;
for (int i = 0, carry = 0; i < (int) v.a.size() || carry; ++i) {
res.a[i] -= carry + (i < (int) v.a.size() ? v.a[i] : 0);
carry = res.a[i] < 0;
if (carry)
res.a[i] += base;
}
res.trim();
return res;
}
return -(v - *this);
}
return *this + (-v);
}
void operator*=(int v) {
if (v < 0)
sign = -sign, v = -v;
for (int i = 0, carry = 0; i < (int) a.size() || carry; ++i) {
if (i == (int) a.size())
a.push_back(0);
long long cur = a[i] * (long long) v + carry;
carry = (int) (cur / base);
a[i] = (int) (cur % base);
//asm("divl %%ecx" : "=a"(carry), "=d"(a[i]) : "A"(cur), "c"(base));
}
trim();
}
bigint operator*(int v) const {
bigint res = *this;
res *= v;
return res;
}
friend pair<bigint, bigint> divmod(const bigint &a1, const bigint &b1) {
int norm = base / (b1.a.back() + 1);
bigint a = a1.abs() * norm;
bigint b = b1.abs() * norm;
bigint q, r;
q.a.resize(a.a.size());
for (int i = a.a.size() - 1; i >= 0; i--) {
r *= base;
r += a.a[i];
int s1 = r.a.size() <= b.a.size() ? 0 : r.a[b.a.size()];
int s2 = r.a.size() <= b.a.size() - 1 ? 0 : r.a[b.a.size() - 1];
int d = ((long long) base * s1 + s2) / b.a.back();
r -= b * d;
while (r < 0)
r += b, --d;
q.a[i] = d;
}
q.sign = a1.sign * b1.sign;
r.sign = a1.sign;
q.trim();
r.trim();
return make_pair(q, r / norm);
}
bigint operator/(const bigint &v) const {
return divmod(*this, v).first;
}
bigint operator%(const bigint &v) const {
return divmod(*this, v).second;
}
void operator/=(int v) {
if (v < 0)
sign = -sign, v = -v;
for (int i = (int) a.size() - 1, rem = 0; i >= 0; --i) {
long long cur = a[i] + rem * (long long) base;
a[i] = (int) (cur / v);
rem = (int) (cur % v);
}
trim();
}
bigint operator/(int v) const {
bigint res = *this;
res /= v;
return res;
}
int operator%(int v) const {
if (v < 0)
v = -v;
int m = 0;
for (int i = a.size() - 1; i >= 0; --i)
m = (a[i] + m * (long long) base) % v;
return m * sign;
}
void operator+=(const bigint &v) {
*this = *this + v;
}
void operator-=(const bigint &v) {
*this = *this - v;
}
void operator*=(const bigint &v) {
*this = *this * v;
}
void operator/=(const bigint &v) {
*this = *this / v;
}
bool operator<(const bigint &v) const {
if (sign != v.sign)
return sign < v.sign;
if (a.size() != v.a.size())
return a.size() * sign < v.a.size() * v.sign;
for (int i = a.size() - 1; i >= 0; i--)
if (a[i] != v.a[i])
return a[i] * sign < v.a[i] * sign;
return false;
}
bool operator>(const bigint &v) const {
return v < *this;
}
bool operator<=(const bigint &v) const {
return !(v < *this);
}
bool operator>=(const bigint &v) const {
return !(*this < v);
}
bool operator==(const bigint &v) const {
return !(*this < v) && !(v < *this);
}
bool operator!=(const bigint &v) const {
return *this < v || v < *this;
}
void trim() {
while (!a.empty() && !a.back())
a.pop_back();
if (a.empty())
sign = 1;
}
bool isZero() const {
return a.empty() || (a.size() == 1 && !a[0]);
}
bigint operator-() const {
bigint res = *this;
res.sign = -sign;
return res;
}
bigint abs() const {
bigint res = *this;
res.sign *= res.sign;
return res;
}
long long longValue() const {
long long res = 0;
for (int i = a.size() - 1; i >= 0; i--)
res = res * base + a[i];
return res * sign;
}
friend bigint gcd(const bigint &a, const bigint &b) {
return b.isZero() ? a : gcd(b, a % b);
}
friend bigint lcm(const bigint &a, const bigint &b) {
return a / gcd(a, b) * b;
}
void read(const string &s) {
sign = 1;
a.clear();
int pos = 0;
while (pos < (int) s.size() && (s[pos] == '-' || s[pos] == '+')) {
if (s[pos] == '-')
sign = -sign;
++pos;
}
for (int i = s.size() - 1; i >= pos; i -= base_digits) {
int x = 0;
for (int j = max(pos, i - base_digits + 1); j <= i; j++)
x = x * 10 + s[j] - '0';
a.push_back(x);
}
trim();
}
friend istream& operator>>(istream &stream, bigint &v) {
string s;
stream >> s;
v.read(s);
return stream;
}
friend ostream& operator<<(ostream &stream, const bigint &v) {
if (v.sign == -1)
stream << '-';
stream << (v.a.empty() ? 0 : v.a.back());
for (int i = (int) v.a.size() - 2; i >= 0; --i)
stream << setw(base_digits) << setfill('0') << v.a[i];
return stream;
}
static vector<int> convert_base(const vector<int> &a, int old_digits, int new_digits) {
vector<long long> p(max(old_digits, new_digits) + 1);
p[0] = 1;
for (int i = 1; i < (int) p.size(); i++)
p[i] = p[i - 1] * 10;
vector<int> res;
long long cur = 0;
int cur_digits = 0;
for (int i = 0; i < (int) a.size(); i++) {
cur += a[i] * p[cur_digits];
cur_digits += old_digits;
while (cur_digits >= new_digits) {
res.push_back(int(cur % p[new_digits]));
cur /= p[new_digits];
cur_digits -= new_digits;
}
}
res.push_back((int) cur);
while (!res.empty() && !res.back())
res.pop_back();
return res;
}
typedef vector<long long> vll;
static vll karatsubaMultiply(const vll &a, const vll &b) {
int n = a.size();
vll res(n + n);
if (n <= 32) {
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++)
res[i + j] += a[i] * b[j];
return res;
}
int k = n >> 1;
vll a1(a.begin(), a.begin() + k);
vll a2(a.begin() + k, a.end());
vll b1(b.begin(), b.begin() + k);
vll b2(b.begin() + k, b.end());
vll a1b1 = karatsubaMultiply(a1, b1);
vll a2b2 = karatsubaMultiply(a2, b2);
for (int i = 0; i < k; i++)
a2[i] += a1[i];
for (int i = 0; i < k; i++)
b2[i] += b1[i];
vll r = karatsubaMultiply(a2, b2);
for (int i = 0; i < (int) a1b1.size(); i++)
r[i] -= a1b1[i];
for (int i = 0; i < (int) a2b2.size(); i++)
r[i] -= a2b2[i];
for (int i = 0; i < (int) r.size(); i++)
res[i + k] += r[i];
for (int i = 0; i < (int) a1b1.size(); i++)
res[i] += a1b1[i];
for (int i = 0; i < (int) a2b2.size(); i++)
res[i + n] += a2b2[i];
return res;
}
bigint operator*(const bigint &v) const {
vector<int> a6 = convert_base(this->a, base_digits, 6);
vector<int> b6 = convert_base(v.a, base_digits, 6);
vll a(a6.begin(), a6.end());
vll b(b6.begin(), b6.end());
while (a.size() < b.size())
a.push_back(0);
while (b.size() < a.size())
b.push_back(0);
while (a.size() & (a.size() - 1))
a.push_back(0), b.push_back(0);
vll c = karatsubaMultiply(a, b);
bigint res;
res.sign = sign * v.sign;
for (int i = 0, carry = 0; i < (int) c.size(); i++) {
long long cur = c[i] + carry;
res.a.push_back((int) (cur % 1000000));
carry = (int) (cur / 1000000);
}
res.a = convert_base(res.a, 6, base_digits);
res.trim();
return res;
}
};
//use : bigint var;
//template for biginit over
int main()
{
bigint var=10909000890789;
cout<<var;
return 0;
}