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
Related
I have one program downloaded from internet and need to get each digit printed out from a three digit number. For example:
Input: 123
Expected Output:
1
2
3
I have 598
Need to Get:
5
9
8
I try using this formula but the problem is when number is with decimal function failed:
FIRST_DIGIT = (number mod 1000) / 100
SECOND_DIGIT = (number mod 100) / 10
THIRD_DIGIT = (number mod 10)
Where number is the above example so here is calulation:
FIRST_DIGIT = (598 mod 1000) / 100 = 5,98 <== FAILED...i need to get 5 but my program shows 0 because i have decimal point
SECOND_DIGIT = (598 mod 100) / 10 = 9,8 <== FAILED...i need to get 9 but my program shows 0 because i have decimal point
THIRD_DIGIT = (598 mod 10) = 8 <== CORRECT...i get from program output number 8 and this digit is correct.
So my question is is there sample or more efficient code that get each digit from number without decimal point? I don't want to use round to round nearest number because sometime it fill failed if number is larger that .5.
Thanks
The simplest solution is to use integer division (\) instead of floating point division (/).
If you replace each one of your examples with the backslash (\) instead of forward slash (/) they will return integer values.
FIRST_DIGIT = (598 mod 1000) \ 100 = 5
SECOND_DIGIT = (598 mod 100) \ 10 = 9
THIRD_DIGIT = (598 mod 10) = 8
You don't have to do any fancy integer calculations as long as you pull it apart from a string:
INPUT X
X$ = STR$(X)
FOR Z = 1 TO LEN(X$)
PRINT MID$(X$, Z, 1)
NEXT
Then, for example, you could act upon each string element:
INPUT X
X$ = STR$(X)
FOR Z = 1 TO LEN(X$)
Q = VAL(MID$(X$, Z, 1))
N = N + 1
PRINT "Digit"; N; " equals"; Q
NEXT
Additionally, you could tear apart the string character by character:
INPUT X
X$ = STR$(X)
FOR Z = 1 TO LEN(X$)
SELECT CASE MID$(X$, Z, 1)
CASE " ", ".", "+", "-", "E", "D"
' special char
CASE ELSE
Q = VAL(MID$(X$, Z, 1))
N = N + 1
PRINT "Digit"; N; " equals"; Q
END SELECT
NEXT
I'm no expert in Basic but looks like you have to convert floating point number to Integer. A quick google search told me that you have to use Int(floating_point_number) to convert float to integer.
So
Int((number mod 100)/ 10)
should probably the one you are looking for.
And, finally, all string elements could be parsed:
INPUT X
X$ = STR$(X)
PRINT X$
FOR Z = 1 TO LEN(X$)
SELECT CASE MID$(X$, Z, 1)
CASE " "
' nul
CASE "E", "D"
Exponent = -1
CASE "."
Decimal = -1
CASE "+"
UnaryPlus = -1
CASE "-"
UnaryNegative = -1
CASE ELSE
Q = VAL(MID$(X$, Z, 1))
N = N + 1
PRINT "Digit"; N; " equals"; Q
END SELECT
NEXT
IF Exponent THEN PRINT "There was an exponent."
IF Decimal THEN PRINT "There was a decimal."
IF UnaryPlus THEN PRINT "There was a plus sign."
IF UnaryNegative THEN PRINT "There was a negative sign."
I'm trying to solve the FizzBuzz game.
I need to check if a number is divisible by 3. So, when we take a number and divide it by 3, we need that operation to have NO REST, or 0 REST.
The solution given to me is this:
def fizzbuzz(number)
if number % 3 == 0
return "Fizz"
end
Why does they propose to use the "%"symbol? Why not the "/"symbol? Or both work as the same?
How should I check if the division has NO REST?
division operator / - gives the quotient of the division whatever the remainder of the division is. So you cannot determine if a number is perfectly divisible (remainder = 0) or not perfectly divisible (with non-zero remainder) using a division operator (/).
10 / 3
#=> 3
modulo operator % - gives the remainder of the division. If perfectly divisible, the output is 0, if not-perfectly divisible the output is non-zero value.
10 % 3
#=> 1
In your case number % 3 == 0 is true only if number is divisible by 3 with 0 remainder (i.e if number passed into the method frizzbuzz is a multiple of 3 like -12, -3, 3, 6, 9, etc )
Ruby has 4 division operators.
divmod returns the division and the remainder
number = 15
number.divmod(7) # => 2, 1
modulo alias % returns the remainder only
number = 15
number % 7 # => 1
number.modulo(7) # => 1
div alias / returns the integer division if both operands are integers, and floating-point division if either operand is a float.
number = 15
number / 7 # => 2
number.div(7) # => 2
1.0 * number / 7 # => 2.142857142857143
fdiv always returns a full precision floating-point division
number = 15
number.fdiv(7) # => 2.142857142857143
% is the Modulus - Divides left hand operand by right hand operand and returns remainder. \ is just the Divider. No Rest means that x % y == 0.
lets take an example to understand better:
number1 = 12
number2 = 13
lets see if number1 and number2 is devisible by 4 ok?
number1 / 4 = 3 and the rest = 0
number2 / 4 = 3 and the rest = 1
so the "/" operation let us know the result of devision operation
and the "%" operation let us know the rest of devision operation
so if we take our examples the number1 is devisible by 3 because
number1 % 3 = 0 ( the rest )
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.
I have a pseudocode which I'm trying to make a detailed analysis, analyze runtime, and asymptotic analysis:
sum = 0
i = 1
while (i ≤ n){
sum = sum + i
i = 2i
}
return sum
My assignment requires that I write the cost/runtime for every line, add these together, and find a Big-Oh notation for the runtime. My analysis looks like this for the moment:
sum = 0 1
long i = 1 1
while (i ≤ n){ log n + 1
sum = sum + i n log n
i = 2i n log n
}
return sum 1
=> 2 n log n + log n + 4 O(n log n)
is this correct ? Also: should I use n^2 on the while loop instead ?
Because of integer arithmetic, the runtime is
O(floor(ln(n))+1) = O(ln(n)).
Let's step through your pseudocode. Consider the case that n = 5.
iteration# i ln(i) n
-------------------------
1 1 0 5
2 2 1 5
3 4 2 5
By inspection we see that
iteration# = ln(i)+1
So in summary:
sum = 0 // O(1)
i = 1 // O(1)
while (i ≤ n) { // O(floor(ln(n))+1)
sum = sum + i // 1 flop + 1 mem op = O(1)
i = 2i // 1 flop + 1 mem op = O(1)
}
return sum // 1 mem op = O(1)
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).