I was solving this codechef problem on Fibonacci numbers. It says number is of 1000 digits then why it is not causing integer overflow in tester's solution when it is scanning the array and storing it in unsigned long long int. I can't understand how solution is working. Below is the problem and tester's solution.
The Head Chef has been playing with Fibonacci numbers for long . He has learnt several tricks related to Fibonacci numbers . Now he wants to test his chefs in the skills .
A fibonacci number is defined by the recurrence :
f(n) = f(n-1) + f(n-2) for n > 2
and f(1) = 0
and f(2) = 1 .
Given a number A , determine if it is a fibonacci number.
Input
The first line of the input contains an integer T denoting the number of test cases. The description of T test cases follows.
The only line of each test case contains a single integer A denoting the number to be checked .
Output
For each test case, output a single line containing "YES" if the given number is a fibonacci number , otherwise output a single line containing "NO" .
Constraints
1 ≤ T ≤ 1000
1 ≤ number of digits in A ≤ 1000
The sum of number of digits in A in all test cases <= 10000.
Example
Input:
3
3
4
5
Output:
YES
NO
YES
**Tester's solution:**
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <set>
#include <cstring>
using namespace std;
int const mx = 6666;
set <unsigned long long> f;
unsigned long long fib[mx + 10];
char s[mx + 1];
int main(){
// freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
fib[0] = 0;
fib[1] = 1;
f.insert(1);
f.insert(0);
int i;
for (i = 2; i <= mx; i++){
fib[i] = fib[i - 1] + fib[i - 2];
f.insert(fib[i]);
}
int tc;
cin>>tc;
while (tc--){
unsigned long long n = 0, ten = 10;
cin>>s;
int len = strlen(s);
for (i = 0; i < len; i++){
char q = s[i];
unsigned long long a = q - '0';
n = n * ten + a;
}
if (f.find(n) == f.end()) printf("NO\n");
else printf("YES\n");
}
return 0;
}
From cplusplus you will see that,
ULLONG_MAX Maximum value for an object of type unsigned long long int is 18446744073709551615 (264-1) or greater.
The actual value depends on the particular system and library implementation, but shall reflect the limits of these types in
the target platform.
Above information is just to let you know its a BIG number. Moreover the cause of not getting overflow is not the limit i mentioned.
Most probably, the input file of judge does not contain any input that can cause an overflow.
And its still possible to set such input even after fulfilling the conditions,
1 ≤ T ≤ 1000
1 ≤ number of digits in A ≤ 1000
The sum of number of digits in A in all test cases <= 10000.
Related
Sometime arc4random() gives negative number also in objective C.
My code is as follow:
Try 1:
long ii = arc4random();
Try 2:
int i = arc4random();
How can I only get positivite random number?
Thank you,
No, it's always positive as it returns an unsigned 32-bit integer (manpage):
u_int32_t arc4random(void);
You are treating it as a signed integer, which is incorrect.
You should use the arc4random_uniform() function. this is the most common random function used.
arc4random_uniform() function
Returns a random number between 0 and the inserted parameter minus 1.
For example arc4random_uniform(3) may return 0, 1 or 2 but not 3.
Example
u_int32_t randomPositiveNo = arc4random_uniform(5) + 1; //to get the range 1 - 5
In one of my java programs I am trying to read a number and then use the golden ratio (1.618034) to find the next smallest fibonacci number its index. For example, if I enter 100000 I should get back "the smallest fibonacci number which is greater than 100000 is the 26th and its value is 121393".
The program should also calculate a fibonacci number by index (case 1 in the code below) which I have coded so far, but I can't figure out how to solve the problem described above (case 2). I have a horrible teacher and I don't really understand what I need to do. I am not asking for the code, just kind of a step by step what I should do for case 2. I can not use recursion. Thank you for any help. I seriously suck at wrapping my head around this.
import java.util.Scanner;
public class Fibonacci {
public static void main(String args[]) {
Scanner scan = new Scanner(System.in);
System.out.println("This is a Fibonacci sequence generator");
System.out.println("Choose what you would like to do");
System.out.println("1. Find the nth Fibonacci number");
System.out.println("2. Find the smallest Fibonacci number that exceeds user given value");
System.out.println("3. Find the two Fibonacci numbers whose ratio is close enough to the golden number");
System.out.print("Enter your choice: ");
int choice = scan.nextInt();
int xPre = 0;
int xCurr = 1;
int xNew = 0;
switch (choice)
{
case 1:
System.out.print("Enter the target index to generate (>1): ");
int index = scan.nextInt();
for (int i = 2; i <= index; i++)
{
xNew = xPre + xCurr;
xPre = xCurr;
xCurr = xNew;
}
System.out.println("The " + index + "th number Fibonacci number is " + xNew);
break;
case 2:
System.out.print("Enter the target value (>1): ");
int value = scan.nextInt();
}
}
}
First, you should understand what this golden ration story is all about. The point is, Fibonacci numbers can be calced recursively, but there's also a formula for the nth Fibonacci number:
φ(n) = [φ^n - (-φ)^(-n)]/√5
where φ = (√5 + 1)/2 is the Golden Ratio (approximately 1.61803). Now, |(-φ)^(-1)| < 1 which means that you can calc φ(n) as the closest integer to φ^n/√5 (unless n = 1).
So, calc √5, calc φ, then learn how to get an integer closest to the value of a real variable and then calc φ(n) using the φ^n/√5 formula (or just use the "main" [φ^n - (-φ)^(-n)]/√5 formula) in a loop and in that loop compare φ(n) with the number that user input. When φ(n) exceeds the user's number, remember n and φ(n).
I want to generate two different random numbers which has 15 digits. How can I do that
Thanks
Most random number generating functions such as arc4random produce only numbers in
the range 0 .. 2^32-1 = 2147483647. For a 15 digit decimal number, you can compute
3 numbers in the range 0 .. 10^5-1 and "concatenate" them:
uint64_t n1 = arc4random_uniform(100000); // 0 .. 99999
uint64_t n2 = arc4random_uniform(100000);
uint64_t n3 = arc4random_uniform(100000);
uint64_t number = ((n1 * 100000ULL) + n2) * 100000ULL + n3; // 0 .. 999999999999999
Or, if you need exactly 15 digits:
uint64_t n1 = 10000 + arc4random_uniform(90000); // 10000 .. 99999
uint64_t n2 = arc4random_uniform(100000); // 0 .. 99999
uint64_t n3 = arc4random_uniform(100000); // 0 .. 99999
uint64_t number = ((n1 * 100000ULL) + n2) * 100000ULL + n3;
Try this::
arc4random() is the standard Objective-C random number generator function. It'll give you a number between zero and... well, more than fifteen! You can generate a number between 0 and 15 (so, 0, 1, 2, ... 15) :
A random number with 6 digits would be:
int number = arc4random_uniform(900000) + 100000;
it will give random numbers from 100000 to 899999.
Hope it Helps!!
using arc4random() functionality you can achieve to generate random numbers.
Here is the link that will give you pretty good idea about arc4random().
hope this will help
Can I set a range of numbers when using arc4random()? For example 50-100 only.
As pointed out in other posts below, it is better to use arc4random_uniform. (When this answer was originally written, arc4random_uniform was not available). Besides avoiding the modulo bias of arc4random() % x, it also avoids a seeding problem with arc4random when used recursively in short timeframes.
arc4random_uniform(4)
will generate 0, 1, 2 or 3. Thus you could use:
arc4random_uniform(51)
and merely add 50 to the result to get a range between 50 & 100 (inclusive).
To expand upon JohnK comment.
It is suggested that you use the following function to return a ranged random number:
arc4random_uniform(51)
which will return a random number in the range 0 to 50.
Then you can add your lower bounds to this like:
arc4random_uniform(51) + 50
which will return a random number in the range 50 to 100.
The reason we use arc4random_uniform(51) over arc4random() % 51 is to avoid the modulo bias. This is highlighted in the man page as follows:
arc4random_uniform(upper_bound) will return a uniformly distributed random number less than upper_bound. arc4random_uniform() is recommended over constructions like ``arc4random() % upper_bound'' as it avoids "modulo bias" when the upper bound is not a power of two.
In short you get a more evenly distributed random number generated.
int fromNumber = 10;
int toNumber = 30;
int randomNumber = (arc4random()%(toNumber-fromNumber))+fromNumber;
Will generate randon number between 10 and 30, i.e. 11,12,13,14......29
You can use this code for generating random values with range:
//range from 50 to 100
int num1 = (arc4random() % 50) + 50; or
int num1 = arc4random_uniform(50) + 50;
//range from 0-100
int num1 = arc4random() % 100; or
int num1 = arc4random_uniform(100);
In Swift you can use this (inspired by answer of #Justyn)
func generateRandomKey(fromRange rangeFrom:Int, toRange rangeTo:Int) -> Int{
let theKey = arc4random_uniform(UInt32(rangeTo - rangeFrom)) + UInt32(rangeFrom)
return Int(theKey)
}
Will always give you a random range Integer.
In many situations 10 thru 30 would mean inclusive, (includes 10 and 30) ...
int fromNumber = 10;
int toNumber = 30;
toNumber ++;
int randomNumber = (arc4random()%(toNumber-fromNumber))+fromNumber;
Notice the difference toNumber - fromNumber is now 21 ... (20+1) which yields the possible results of 0 thru 20 (inclusive) which when added to fromNumber (10) results in 10 thru 30 (inclusive).
Let's say i have this bit field value: 10101001
How would i test if any other value differs in any n bits. Without considering
the positions?
Example:
10101001
10101011 --> 1 bit different
10101001
10111001 --> 1 bit different
10101001
01101001 --> 2 bits different
10101001
00101011 --> 2 bits different
I need to make a lot of this comparisons so i'm primarily looking for perfomance but any
hint is very welcome.
Take the XOR of the two fields and do a population count of the result.
if you XOR the 2 values together, you are left only with the bits that are different.
You then only need to count the bits which are still 1 and you have your answer
in c:
unsigned char val1=12;
unsigned char val2=123;
unsigned char xored = val1 ^ val2;
int i;
int numBits=0;
for(i=0; i<8; i++)
{
if(xored&1) numBits++;
xored>>=1;
}
although there are probably faster ways to count the bits in a byte
(you could for instance use a lookuptable for 256 values)
Just like everybody else said, use XOR to determine what's different and then use one of these algorithms to count.
This gets the bit difference between the values and counts the bits three at a time:
public static int BitDifference(int a, int b) {
int cnt = 0, bits = a ^ b;
while (bits != 0) {
cnt += (0xE994 >> ((bits & 7) << 1)) & 3;
bits >>= 3;
}
return cnt;
}
XOR the numbers, then the problem becomes a matter of counting the 1s in the result.
In Java:
Integer.bitCount(a ^ b)
Comparison is performed with XOR, as others already answered.
counting can be performed in several ways:
shift left and addition.
lookup in a table.
logic formulas that you can find with Karnaugh maps.