This question already has answers here:
Why does rand() yield the same sequence of numbers on every run?
(7 answers)
Closed 5 years ago.
I created a random number generator (between 100 and 1) for an Array but each time I run it it outputs the same 10 numbers between 1 and 100.
for (int x=0; x < 10; x++) {
arr[x]=rand() % 100 + 1;
The Loop runs 10 times because the array contains 10 integers. Every time I run this though, it outputs: 42, 68, 35, 1, 70, 25, 79, 59, 63, 65.
It Outputs from this line of code
for (int z=0; z<10;z++) {
cout << arr[z]<<", ";
}
Can anyone see why this is happening?
When using rand() it will generate the same sequence from the given input to srand(). Another words it will typically use the same generated numbers from the initial seed value. These functions are considered to be frowned upon in their use and some even want them to be marked as deprecated in favor of the modern std pseudo-random number generators library.
To generate random numbers try using the library from the <random> header instead.
You can find the documentations here: cppreference::random
Related
This question already has answers here:
C++ random number same sequence every time
(2 answers)
Random seed at runtime
(3 answers)
Closed 1 year ago.
I'm using rand() in a loop to generate random numbers every time till the loop is complete, but it always gives the same number, what am I doing wrong?
bool PlayGame(int Difficulty, bool bComplete)
{
int CodeA =rand() % Difficulty + Difficulty;
int CodeB =rand() % Difficulty + Difficulty;
int CodeC =rand() % Difficulty + Difficulty;
You can use current time as seed for random generator by setting srand(time(0)); at start
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
// Driver program
int main(void)
{
// This program will create different sequence of
// random numbers on every program run
// Use current time as seed for random generator
srand(time(0));
for(int i = 0; i<4; i++)
printf(" %d ", rand());
return 0;
}
Output 1:
453 1432 325 89
Output 2:
8976 21234 45 8975
Output n:
563 9873 12321 24132
Ref.
If random numbers are generated with rand() without first calling srand(), your program will create the same sequence of numbers each time it runs.
The srand() function sets the starting point for producing a series of pseudo-random integers. If srand() is not called, the rand() seed is set as if srand(1)
so, set srand(time(0)); at start of the program
This question already has answers here:
What is the behavior of integer division?
(6 answers)
Integer division always zero [duplicate]
(1 answer)
Random number c++ in some range [duplicate]
(6 answers)
Closed 3 years ago.
I need to produces numbers between 0 and a max (seen in code as assetMax). In the code, the rand()/RAND_MAX always produces 0 and I cannot seem to figure out why. I use the rand() function immediately before it to produce values in a range and it works completely fine. However, here it does not.
I have tried to switch the order of the variables, create the random number in a separate double before multiplying the two, and the header.
void cPortfolio::randomize(cProblem &portfolioProblem) {
int assetCount = 6 * rand() / RAND_MAX + (portfolioProblem.assetMax-8); //this line works as expected
int test;
for (int i = 0; i < assetCount; i++) {
double num = rand() / RAND_MAX; //this always produces 0.0000
int test = num * (portfolioProblem.assetNum); } `} //cannot format these correctly please ignore the brackets
This question already has answers here:
Generate random numbers uniformly over an entire range
(20 answers)
rand() returns same values when called within a single function
(5 answers)
Closed 4 years ago.
We will modify part of the existing menu logic to generate an ID for the user. This user ID will be of the type int, and will be 5 digits in length. The values are 00001 – 99999, and should appear with all 5 digits (leading zeroes) on the display upon generating them. Do not change the data type from that of an int. Check out the setfill manipulator as a possible helper for this. Setfill allows you to set a fill character to a value using an input parameter, such as ‘0’ for leading zeroes. Researching it, you’ll see that you will need to reset the fill character after use too, or you will be printing a lot of zeroes on the display! Your program also should guarantee an absolutely unique user ID has been generated for each new activity that is added. Think about how this works...
Currently I've been trying to get the following code to work
srand(time(0));
cout << setfill('0') << setw(5) << rand() %99999 << endl;
Problem is that this doesn't seem random at all (it's just slowly counting up based on the computers internal clock right?) and the first digit is always zero. Like the instructions say it should be between 00001 and 99999.
EDIT: I appreciate the different solutions, but most of them are more advanced than what I'm supposed to be using for this assignment. I'm fairly sure srand() and rand() is what I should be using.
Okay, so it seems you must use the rand() function to generate a value between 1-99999. So with that in mind, the following code should generate random values in the required range:
#include <iomanip>
#include <iostream>
const int randID()
{
return 1 + (std::rand() % (99999));
}
int main()
{
for(int i = 0; i < 1000; ++i)
std::cout << std::setfill('0') << std::setw(5) << randID() << '\n';
return 0;
}
For me it prints, for example:
01886
21975
01072
11334
22868
26154
14296
32169
20826
09677
15630
28651
Which should satisfy your requirement of 0-padded values between 1-99999. Also, as mentioned in the comments. Do look into the <random> for your random number needs outside this assignment as it generates far superior random numbers, and offers way more generators, distributions and better seeding.
This question already has answers here:
How do I scale down numbers from rand()?
(9 answers)
Closed 8 years ago.
Hi. Is there any way to set the size of random numbers?( in random number generator "rand()")
For example I want to generate 10 digits random numbers.
and one more question, how can i set random function to generate numbers between 0 and 1 (for example 0100110110) ?
Im not sure about setting the size of the numbers. However I dont think it would be possible to get each digit to produce just a 0 or 1.
What you can do however is something like below:
ostringstream 10digitNumber;
for(int i = 0 ; i < 10 ; i ++){
v1 = rand() % 2;// generate o or 1
10digitNumber<< v1;// build up a string of 1 and 0
}
int real10DigitNumber = static_cast<int>10digitNumber); // typecast to integer
Please forgive me if my syntax isn't 100 %. Its being awhile since I used c++.
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Project Euler Problem 12 - C++
The sequence of triangle numbers is generated by adding the natural numbers. Hence, the 7th
triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28.
The first ten terms would be:
1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...
Let us list the factors of the first seven triangle numbers:
1: 1
3: 1,3
6: 1,2,3,6
10: 1,2,5,10
15: 1,3,5,15
21: 1,3,7,21
28: 1,2,4,7,14,28
We can see that 28 is the first triangle number to have over five divisors. What is the value of the
first triangle number to have over one hundred divisors?
You've only copied the problem description! What's your problem with the problem? You have to state that.
The question is posed "what is the 1st triangle number to have over one hundred divisors?" Simply iterate over the triangle numbers, finding out how many factors each one has. When you find one with >100 factors, you're done.
for each whole number 'n' from 1 -> +INF
let tn = triangleNumber(n);
let nf = numFactors(tn);
if (nf > 100)
print tn " has " nf " factors.\n";
return;
Firstly try do it by yourself. If you are not able to get your answer then understand this code.Try to understand the problem and then try to impliment it by yourself. Firstly you have to check untill your divisor is over 100 so there would be one while loop.Inside that while you have to create that sequence of triangle i.e sum of consecutive numbers (1+2+3+4+5+6+7). And then use counter and increment it to find out the number of divisor of the sum.
#include <iostream>
#include <stdlib.h>
using namespace std;
int main(){
int div=0,sum=0,num,i=1,chk=0,a;
cout<<"enter the number of divisors"<<endl;
cin>>a;
while(div<=a)
{div=0;
sum=sum+i;
for(int j=1;j<=sum;j++)
{if(sum%j==0)
div++;
}
chk++;
i++;
}
cout<<"Value of first triangle number value is "<<sum<<endl;
cout<<"Value of triangle number is "<<chk<<endl;
system("PAUSE");
return 0;
}