I have a float that has a value like this: 0.012447, I want to rescale the value I have between 0.0 - 1.0 and use it to display a percentage, but I would like to only display even values, like ... 12%, 14%, 16%, ...
Something like this:
float percent = ....
int value = percent * 100;
if(value == 12 || value == 14 || value == 16 .....){
printf("Result: = %f\n", percent);
}
If there is an easy way to do it without using the percent * 10 and multiple ifs i would like examples
As a complement to #salcc's answer, this is probably close to what you ask for; conditional printing of value if it's even.
#include <cmath>
#include <iostream>
int main() {
for(float percent = 0.f; percent <= 1.f; percent += 1.f / 47.f) {
int value = static_cast<int>(std::round(percent * 100.f));
if(value % 2 == 0) { // Is it even? If so, print it.
std::cout << "Result: = " << value << "\n";
}
}
}
What's really bad with this approach is that it may create gaps in the output. The above creates this output:
Result: = 0
Result: = 2
Result: = 4
Result: = 6
Result: = 26
Result: = 28
Result: = 30
Result: = 32
Result: = 34
Result: = 36
Result: = 38
Result: = 40
Result: = 60
Result: = 62
Result: = 64
Result: = 66
Result: = 68
Result: = 70
Result: = 72
Result: = 74
Result: = 94
Result: = 96
Result: = 98
Result: = 100
To round to the nearest even number you can divide the number by two, round the result, and then multiply it by two. You can do that and convert it to a percentage by just doing:
printf("Result: %g%%\n", round(percent * 100 / 2.0) * 2);
To use the round() function, you have to put #include <cmath> at the top of the file.
Demo:
float percent = 0.129;
printf("Result: %g%%\n", round(percent * 100 / 2.0) * 2);
Output:
Result: 12%
A number is called a stepping number if all adjacent digits in the number have an absolute difference of 1.
Examples of stepping numbers :- 0,1,2,3,4,5,6,7,8,9,10,12,21,23,...
I have to generate stepping numbers upto a given number N. The numbers generated should be in order.
I used the simple method of moving over all the numbers upto N and checking if it is stepping number or not. My teacher told me it is brute force and will take more time. Now, I have to optimize my approach.
Any suggestions.
Stepping numbers can be generated using Breadth First Search like approach.
Example to find all the stepping numbers from 0 to N
-> 0 is a stepping Number and it is in the range
so display it.
-> 1 is a Stepping Number, find neighbors of 1 i.e.,
10 and 12 and push them into the queue
How to get 10 and 12?
Here U is 1 and last Digit is also 1
V = 10 + 0 = 10 ( Adding lastDigit - 1 )
V = 10 + 2 = 12 ( Adding lastDigit + 1 )
Then do the same for 10 and 12 this will result into
101, 123, 121 but these Numbers are out of range.
Now any number transformed from 10 and 12 will result
into a number greater than 21 so no need to explore
their neighbors.
-> 2 is a Stepping Number, find neighbors of 2 i.e.
21, 23.
-> generate stepping numbers till N.
The other stepping numbers will be 3, 4, 5, 6, 7, 8, 9.
C++ code to do generate stepping numbers in a given range:
#include<bits/stdc++.h>
using namespace std;
// Prints all stepping numbers reachable from num
// and in range [n, m]
void bfs(int n, int m)
{
// Queue will contain all the stepping Numbers
queue<int> q;
for (int i = 0 ; i <= 9 ; i++)
q.push(i);
while (!q.empty())
{
// Get the front element and pop from the queue
int stepNum = q.front();
q.pop();
// If the Stepping Number is in the range
// [n, m] then display
if (stepNum <= m && stepNum >= n)
cout << stepNum << " ";
// If Stepping Number is 0 or greater than m,
// need to explore the neighbors
if (stepNum == 0 || stepNum > m)
continue;
// Get the last digit of the currently visited
// Stepping Number
int lastDigit = stepNum % 10;
// There can be 2 cases either digit to be
// appended is lastDigit + 1 or lastDigit - 1
int stepNumA = stepNum * 10 + (lastDigit- 1);
int stepNumB = stepNum * 10 + (lastDigit + 1);
// If lastDigit is 0 then only possible digit
// after 0 can be 1 for a Stepping Number
if (lastDigit == 0)
q.push(stepNumB);
//If lastDigit is 9 then only possible
//digit after 9 can be 8 for a Stepping
//Number
else if (lastDigit == 9)
q.push(stepNumA);
else
{
q.push(stepNumA);
q.push(stepNumB);
}
}
}
//Driver program to test above function
int main()
{
int n = 0, m = 99;
// Display Stepping Numbers in the
// range [n,m]
bfs(n,m);
return 0;
}
Visit this link.
The mentioned link has both BFS and DFS approach.
It will provide you with explaination and code in different languages for the above problem.
We also can use simple rules to move to the next stepping number and generate them in order to avoid storing "parents".
C.f. OEIS sequence
#include <iostream>
int next_stepping(int n) {
int left = n / 10;
if (left == 0)
return (n + 1); // 6=>7
int last = n % 10;
int leftlast = left % 10;
if (leftlast - last == 1 & last < 8)
return (n + 2); // 32=>34
int nxt = next_stepping(left);
int nxtlast = nxt % 10;
if (nxtlast == 0)
return (nxt * 10 + 1); // to get 101
return (nxt * 10 + nxtlast - 1); //to get 121
}
int main()
{
int t = 0;
for (int i = 1; i < 126; i++, t = next_stepping(t)) {
std::cout << t << "\t";
if (i % 10 == 0)
std::cout << "\n";
}
}
0 1 2 3 4 5 6 7 8 9
10 12 21 23 32 34 43 45 54 56
65 67 76 78 87 89 98 101 121 123
210 212 232 234 321 323 343 345 432 434
454 456 543 545 565 567 654 656 676 678
765 767 787 789 876 878 898 987 989 1010
1012 1210 1212 1232 1234 2101 2121 2123 2321 2323
2343 2345 3210 3212 3232 3234 3432 3434 3454 3456
4321 4323 4343 4345 4543 4545 4565 4567 5432 5434
5454 5456 5654 5656 5676 5678 6543 6545 6565 6567
6765 6767 6787 6789 7654 7656 7676 7678 7876 7878
7898 8765 8767 8787 8789 8987 8989 9876 9878 9898
10101 10121 10123 12101 12121
def steppingNumbers(self, n, m):
def _solve(v):
if v>m: return 0
ans = 1 if n<=v<=m else 0
last = v%10
if last > 0: ans += _solve(v*10 + last-1)
if last < 9: ans += _solve(v*10 + last+1)
return ans
ans = 0 if n>0 else 1
for i in range(1, 10):
ans += _solve(i)
return ans
So I'm trying to use a modified Bellman Ford algorithm to find the shortest path from the starting vertex to the ending vertex but I cannot go over a certain distance. So given a graph with edges:
0 1 100 30
0 4 125 50
1 2 50 250
1 2 150 50
4 2 100 40
2 3 90 60
4 3 125 150
Where the each line represents an edge and the first value is the starting vertex, the second value is the end vertex, the third is cost and the fourth is the distance.
With the code I have now when I try to find the cheapest path from 0 to 3 without going over 140 it yields 0 (default when no path is found) instead of 340 (the cost of the cheapest path). Any suggestions on how to alter my code.
Just gonna copy the code down below because this site is not letting me do anything else.
static void BellmanFord(struct Graph *graph, int source, int ending, int max){
int edges = graph->edgeCount;
int vertices = graph->verticesCount;
int* money = (int*)malloc(sizeof(int) * vertices);
int* distance = (int*)malloc(sizeof(int) * vertices);
for (int I = 0; I < vertices; I++){
distance[I] = INT_MAX;
money[I] = INT_MAX;
}
distance[source] = 0;
money[source] = 0;
for (int I = 1; I <= vertices - 1; ++I){
for int j = 0; j < edges; ++j){
int u = graph->edge[j].Source;
int v = graph->edge[j].Destination;
int Cost = graph->edge[j].cost;
int Duration = graph->edge[j].duration;
if ((money[u] != INT_MAX) && (money[u] + Cost < money[v])){
if (distance[u] + Duration <= max){
money[v] = money[u] + Cost;
distance[v] = distance[u] + Duration;
}
}
}
}
if (money[ending] == INT_MAX) cout << "0" << endl;
else cout << money[ending] << endl;
}
Please help! This is probably not that hard but finals are stressing me out
This problem, known as the "constrained shortest path" problem, is much harder to solve than this. The algorithm you provided does not solve it, it only might catch the solution, only by luck, according to the graph's structure.
When this algorithm is applied on the graph you provide, with max-distance = 140, it fails to find the solution, which is 0-->1-->2-->3 (using the edge 1 2 150 50) with total cost of 340 and a distance of 140.
We can observe the reason of failure by logging the updates to the nodes whenever they are updated, and here is the result:
from node toNode newCost newDistance
0 1 100 30
0 4 125 50
1 2 250 80
4 2 225 90
Here the algorithm is stuck and cannot go further, since any progress from this point will lead to paths that exceed the max distance (of 140). As you see, node 2 has found the path 0-->4--2 which is the lowest-cost from node 0 while respecting the max-distance constraint. But now, any progress from 2 to 3 will exceed the distance of 140 (it will be 150, because 2->3 has a distance of 60.)
Running again with max-distance=150 will find the path 0-->4->3 with cost 315 and distance 150.
from node toNode newCost newDistance
0 1 100 30
0 4 125 50
1 2 250 80
4 2 225 90
2 3 315 150
Obviously this is not the minimum cost path that respects the constraint of distance; the correct should be the same (that it failed to find) in the first example. This again proves the failure of the algorithm; this time it gives a solution but which is not the optimal one.
In conclusion, this is not a programming mistake or bug in the code, it is simply that the algorithm is not adequate to the stated problem.
Okay so right before the
if (money[ending] == INT_MAX) cout << "0" << endl;
I added some code that made it work but I'm wondering will this work for every case or does it need to be altered a little.
if (money[ending] == INT_MAX){
for (int j = 0; j < edges; ++j){
int u = graph->edge[j].Source;
int v = graph->edge[j].Destination;
int Cost = graph->edge[j].cost;
int Duration = graph->edge[j].duration;
if ((distance[u] != INT_MAX) && (distance[u] +Duration < distance[v])){
if (distance[u] + Duration <= max){
money[v] = money[u] + Cost;
distance[v] = distance[u] + Duration;
}
}
}
}
This question already has answers here:
Get `n` random values between 2 numbers having average `x`
(5 answers)
Closed 6 years ago.
Problem: Getting a set of random numbers between two values that will have a certain mean value.
Let say we getting n number of random number where the number will be between 1 and 100. We have a mean of 25.
My first approach is to have 2 modes where we have aboveMean and belowMean where the first random number is the initial range 1 and 100. Every subsequent number will check the total sum. If the total sum is above the mean, we go to case aboveMean which is then get a random number between 1 and 25. If the total sum is below the mean, we do case belowMean then get a random number between 26 and 100.
I need some idea on how to approach this problem beside the crude get a random number to add it to the total then get the average. If it above the mean in question, we get a random number below the mean and so forth. While it does work, doesn't seem to be the best method.
I'm guessing I should brush up on probability to approach this random number generator.
Let us divide the range into left and right portions. Use a value from the portion at a frequency corresponding to the relative width of the other half.
int Leruce_rand(int min, int mean, int max) {
int r = rand()%(max - min + 1);
if (r < mean) {
// find number in right half
return rand()%(max - mean + 1) + mean;
} else {
// find number in left half
return rand()%(mean - min) + min;
}
Assumes mean is part of the right half. This quick solution likely has small bias.
Given OP's values, roughly, the average of the left half is 12.5 and called 75% of the time. Average of the right is 62.5 called 25% of the time: average 25.
This approach differs from OP's which "Every subsequent number will check the total sum. If the total sum is above the mean, we go to case aboveMean which is then get a random number between 1 and 25." As that absolutely prevents a set of occurrences above or below the mean. With RNG, the value generated should not be biased on the history of previous generated values.
There's literally an infinite number of ways to achieve this. For instance, generate 3 random numbers between 1 and 100 (std::uniform_int_distribution) and take the minimum of those (std::min(a,b,c)).
Obviously, for a mean of 75 you'll need to pick the maximum of 3 numbers.
The benefit of this method is that each outcome is independent of the previous ones. It's completely random.
Take some good distribution and use it. Say, Binomial distribution. Use B(99,24/99),
so sampled values are in the range 0...99, with parameter p equal to 24/99.
So if you have routine which sample from B, then all you need is to add 1
to be in he range 1...100
Mean value for binomial would be p*n, in this case equal to 24. Because you're adding 1, your mean value would be 25 as required. C++11 has binomial RNG in the
standard library
Some code (not tested)
#include <iostream>
#include <random>
int main() {
std::default_random_engine generator;
std::binomial_distribution<int> distribution(99, double(24)/double(99));
for (int i=0; i != 1000; ++i) {
int number = distribution(generator) + 1;
std::cout << number << std::endl;
}
return 0;
}
Assume a fair random(a,b) function (this question should not be about which random function is better) then simply just restrcting ithe ranges that is piced from should be a good start, like;
const int desiredCount = 16;
const int deiredMean = 25;
int sumValues = random(a,b);
int count = 1;
while (count < desriredCount - 1) {
int mean = sumValue/count;
int nextValue = 0;
if (mean < desiredMean) // Too small, reduce probablity of smaller numbers
nextValue = random(a+(desiredMean-mean)/(desriredCount-count),b);
else //too large, reduce probability of larger numbers
nextValue = random(a,b-(mean-desiredMean)/(desriredCount-count));
sumValue += nextValue;
count += 1;
}
int lastValue = desiredMean*desriredCount - sumValue/count;
sumValue += lastValue;
count += 1;
Note: The above is not tested, and my thinking is that the trimming of the upper and lower bound may not be sufficently aggressive to do the trick, but I hope that i will get you going.
Some boundary conditions, such as if you only want 2 numbers and a means of 25 from numbers between 0 and 100, the initial random number cannot be larger than 50, since that makes it impossible to pick the second (last) number -- so if you want the algo to give you exact mean values under all circumstances, then a bit more tweaking is needed.
OP's wants a set of numbers meeting certain criteria.
Consider generating all possible sets of n numbers in the range [min max] and then eliminating all sets but those with the desired mean. Now randomly select one of those sets. This would meet OP's goal and IMO would pass fair randomness tests. Yet this direct approach is potentially a huge task.
Alternatively, randomly generate lots of sets until one is found that meets the mean test.
The below meets OP's requirement of a specified mean without directly biasing the random numbers selected. Certainly not an efficient method when the desired mean is far from the min/max average.
#include <stdio.h>
#include <stdlib.h>
void L_set(int *set, size_t n, int min, int mean, int max) {
assert(n > 0);
assert(min >= 0);
assert(mean >= min);
assert(max >= mean);
size_t i;
long long diff;
long long sum_target = n;
unsigned long long loop = 0;
sum_target *= mean;
int range = max - min + 1;
do {
loop++;
long long sum = 0;
for (i = 1; i < n; i++) {
set[i] = rand() % range + min;
sum += set[i];
}
diff = sum_target - sum; // What does the final number need to be?
} while (diff < min || diff > max);
set[0] = (int) diff;
printf("n:%zu min:%d mean:%2d max:%3d loop:%6llu {", n, min, mean, max, loop);
for (i = 0; i < n; i++) {
printf("%3d,", set[i]);
}
printf("}\n");
fflush(stdout);
}
int main(void) {
int set[1000];
L_set(set, 10, 1, 2, 4);
L_set(set, 16, 1, 50, 100);
L_set(set, 16, 1, 25, 100);
L_set(set, 16, 1, 20, 100);
return 0;
}
Output
n:10 min:1 mean: 2 max: 4 loop: 1 { 4, 2, 4, 3, 2, 1, 1, 1, 1, 1,}
n:16 min:1 mean:50 max:100 loop: 2 { 45, 81, 24, 50, 93, 65, 70, 52, 28, 91, 25, 36, 21, 45, 11, 63,}
n:16 min:1 mean:25 max:100 loop: 3257 { 52, 1, 15, 70, 66, 30, 1, 4, 26, 1, 16, 4, 48, 42, 19, 5,}
n:16 min:1 mean:20 max:100 loop:192974 { 24, 10, 13, 3, 3, 53, 22, 12, 29, 1, 7, 6, 90, 11, 20, 16,}
you have to go into some probabilities theory. there are a lot of methods to judge on random sequence. for example if you lower the deviation you will get triangle-looking-on-a-graph sequence, which can in the end be proven not trully random. so there is not really much choice than getting random generator and discarding the sequences you don't like.
EDIT: this generates numbers in the range 1..100 with a theoretical mean of 25.25. It does this by using a random modulus in the range 1..100. Note that the required mean is 25, which is not exactly a quarter of the range 1..100.
OP wanted a way of varying the next number chosen according to whether the mean is less than or more than 25, but that lends some predictabilty - if the mean is more than 25 then you know the next "random" number will be less than 25.
The random calculation in the code is a very simple one line.
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#define RUNS 10000000
#define MAXN 100
int main() {
int n, i, sum = 0, min = MAXN, max = 0;
int freq[MAXN+1] = {0};
srand((unsigned)time(NULL));
for(i = 0; i < RUNS; i++) {
n = 1 + rand() % (1 + rand() % 100); // average modulus is (1 + MAX) / 2
if(max < n) {
max = n; // check that whole range is picked
}
if(min > n) {
min = n;
}
freq[n]++; // keep a tally
sum += n;
}
// show statistis
printf("Mean = %f, min = %d, max = %d\n", (double)sum / RUNS, min, max);
for(n = MAXN; n > 0; n--) {
printf("%3d ", n);
for(i = (freq[n] + 5000) / 10000; i > 0; i--) {
printf("|");
}
printf("\n");
}
return 0;
}
Program output showing distribution / 10000:
Mean = 25.728128, min = 1, max = 100
100
99
98
97
96 |
95 |
94 |
93 |
92 |
91 |
90 |
89 |
88 |
87 |
86 ||
85 ||
84 ||
83 ||
82 ||
81 ||
80 ||
79 ||
78 |||
77 |||
76 |||
75 |||
74 |||
73 |||
72 |||
71 ||||
70 ||||
69 ||||
68 ||||
67 ||||
66 ||||
65 ||||
64 |||||
63 |||||
62 |||||
61 |||||
60 |||||
59 |||||
58 ||||||
57 ||||||
56 ||||||
55 ||||||
54 ||||||
53 ||||||
52 |||||||
51 |||||||
50 |||||||
49 |||||||
48 |||||||
47 ||||||||
46 ||||||||
45 ||||||||
44 ||||||||
43 |||||||||
42 |||||||||
41 |||||||||
40 |||||||||
39 ||||||||||
38 ||||||||||
37 ||||||||||
36 ||||||||||
35 |||||||||||
34 |||||||||||
33 |||||||||||
32 ||||||||||||
31 ||||||||||||
30 ||||||||||||
29 |||||||||||||
28 |||||||||||||
27 |||||||||||||
26 ||||||||||||||
25 ||||||||||||||
24 ||||||||||||||
23 |||||||||||||||
22 |||||||||||||||
21 ||||||||||||||||
20 ||||||||||||||||
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18 |||||||||||||||||
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13 |||||||||||||||||||||
12 ||||||||||||||||||||||
11 |||||||||||||||||||||||
10 ||||||||||||||||||||||||
9 |||||||||||||||||||||||||
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7 |||||||||||||||||||||||||||
6 |||||||||||||||||||||||||||||
5 |||||||||||||||||||||||||||||||
4 |||||||||||||||||||||||||||||||||
3 |||||||||||||||||||||||||||||||||||||
2 ||||||||||||||||||||||||||||||||||||||||||
1 ||||||||||||||||||||||||||||||||||||||||||||||||||||
OP did not state what kind of distribution was wanted, for example two straight lines pivoting at 25, or perhaps equal distribution each side of 25. However this solution is very simple to implement.
I'm making a program where I have a list of 34 cities and I am wanting to give each of these cities a probability of being chosen.
So I have:
vector<float> vec;
int s;
cin >> s;
srand(s);
for (int k=0; k < 34; k++)
{
float p1= (float)rand()/(float)((unsigned)RAND_MAX+1);
vec.push_back(p1);
}
So that gives each city a probability. The problem I am now having is I want to then do a random number generator that will choose two of these cities. So, for example city1 will have a 5%, city2 a 2%, city3, a 3%, etc. How can I randomly choose two of these cities based off the probabilities given?
I did this in genetic algorithm.
for your cities consider a line of 10 units.
now from 0-5 units on line are city1 6-7 for city2 and 8-9 for city3.
now choose a number at random from 0-9.
and found out in which cities range it comes in.
At first glance my solution will be :
Create a number equals to all city's probability
Create a random number, with max random number is equal to the previous number
Take the random number, and go throught your city vector and take the one who is corresponding.
Example :
City 1 : 5%
City 2 : 8%
City 3 : 2%
City 4 : 5%
City 5 : 12%
Create a number -> Number a = 32 (5+8+2+5+12)
Generate a number with : 1 <= number
Assume that the number is equal to 12
City 1 is choose if : 1 <= number <= 5 (So not)
City 2 is choose if : 6 <= number <= 13 (So yes)
City 2 is choose.
If you have any questions about that, you are welcome :)
Edit :
Well i will give you some more explaination.
Take this code :
for (int k=0; k < 10; k++)
{
float p1= (float)rand()/(float)((unsigned)RAND_MAX+1);
vec.push_back(p1);
}
Assume now that vec contain the following informations :
5
3
8
5
12
14
8
5
6
18
With each number correspond to the probability to choose a city.
5 -> 5% probability to choose (City1)
3 -> 3% probability to choose (City2)
8 -> 8% probability to choose (City3)
... etc
Now i will give you some code and i will explain it :
int nbReference = 0;
for (auto it = vec.begin() ; it != vec.end() ; ++it)
{
nbReference += *it;
}
nbReference = rand(nbReference);
int i = 0;
int cityProbability = 0;
for (auto it = vec.begin() ; it != vec.end() ; ++it)
{
cityProbability = *it;
if ((i + cityProbability) > nbReference)
{
return (i + 1);
}
i += cityProbability;
}
First i create a number equals to the addition of all city's probability
int nbReference = 0;
for (auto it = vec.begin() ; it != vec.end() ; ++it)
{
nbReference += *it;
}
Second, i generate a number that is respect the following range -> 0 < nbReference
Third, i create a loop that take all city one by one and quit when we got right city.
How does we know when a city is good?
Let's take an example!
With our previous probability
5 3 8 5 12 14 8 5 6 18
NbReference is equals to (5+3+8+5+12+14+8+5+6+18) so 84
To each city we are going to put a range equals to his probability plus all of previous city's probability. Let me show you :
5 -> Range 0 to 4 (0 to 4 = 5 ---> 5%)
3 -> Range 5 to 8 (5 to 8 = 3 ---> 3%)
8 -> Range 9 to 17
5 -> Range 18 to 22
... etc
If the number that we create here
nbReference = rand(nbReference);
Is in a city range, so that city is choosed.
Example : If the number is 16, city3 is choosed!
5 -> Range 0 to 4 Number is 16 so NOPE
3 -> Range 5 to 8 Number is 16 so NOPE
8 -> Range 9 to 17 Number is 16 so YES!
5 -> Range 18 to 22
... etc
Does is this helpfull? :)
Any questions? You are welcome
Maybe this code can help you (follows partially WhozCraig advice)
#include <iostream>
#include <random>
#include <algorithm>
int main(int argc, const char * argv[])
{
using namespace std::chrono;
system_clock::time_point tp = system_clock::now();
system_clock::duration dtn = tp.time_since_epoch();
std::default_random_engine generator(static_cast<int>(dtn.count()));
//Generate 34 cities
std::uniform_real_distribution<double> gen_distribution(0,1);
auto getProb = std::bind ( gen_distribution, generator );
std::vector<double> citiesProb;
double probSum(0.0);
double cityProb(0.0);
for (int k=0; k < 34; k++)
{
cityProb = getProb();
probSum += cityProb;
citiesProb.push_back(cityProb);
}
//Pick 7 cities
std::uniform_real_distribution<double> pick_distribution(0,probSum);
auto pickCity = std::bind ( pick_distribution, generator );
double chooseCity;
double probBasket;
for (int k=0; k < 7; ++k)
{
probBasket = 0.0;
chooseCity = pickCity();
for (int i = 0; i < citiesProb.size(); ++i)
{
if (chooseCity >= probBasket && chooseCity < probBasket + citiesProb[i])
{
std::cout << "City with index " << i << " picked" << std::endl;
}
probBasket += citiesProb[i];
}
}
return 0;
}
How it works:
city1 5%(0.05), city2 25%(0.25), city3 8%(0.08), city4 10%(0.1)
then
probSum = 0.05 + 0.25 + 0.08 + 0.1 = 0.48
then choose a number between 0 and 0.48 (named pickProb) and
if pickProb is between 0 and 0.05 pick city1 (prob = 0.05/0.48 = 10%)
if pickProb is between 0.05 and 0.30 pick city2 (prob = 0.25/0.48 = 52%)
if pickProb is between 0.30 and 0.38 pick city3 (prob = 0.08/0.48 = 16%)
if pickProb is between 0.38 and 0.48 pick city4 (prob = 0.1/0.48 = 20%)
if probSum = 1.0 then city1 is picked with probability 5%, city2 is picked with probability 25% and so on.