Strong number is the number that the sum of the factorial of its digits is equal to number itself.
For example: 145, since
1! + 4! + 5! = 1 + 24 + 120 = 145
Here is my code, It passes most of the test except one test
#include <string>
using namespace std;
string strong_num (int number )
{
int sum = 0;
while(number != 0) {
int last = number % 10;
number /= 10;
sum+= last * (last-1);
}
if(sum == number)
return "STRONG!!!!";
else
return "Not Strong !!";
}
What is wrong with my code?
I'm surprised you're passing any test cases at all. For one thing, you are destroying number before you compare it to sum, and for another your logic is flawed.
Try this:
int factorial (int x)
{
int result = 1;
while (x > 1)
{
result *= x;
x--;
}
return result;
}
string strong_num (int number)
{
int sum = 0;
int x = number;
while (x != 0) {
int digit = x % 10;
sum += factorial (digit);
x /= 10;
}
if (sum == number)
return "STRONG!!!!";
else
return "Not Strong !!";
}
Live demo
Replace int by long long to be able to test larger numbers.
There are two problems:
first - you are changing the value of number before comparing it to sum,
second - the thing you used last * (last-1) is not a definition of factorial, the definition of factorial is factorial(x) = 1 * 2 * 3 * ... * x
int factorial (int x) {
if(x < 2) return 1;
return x * factorial(x - 1);
}
string strong_num (int number)
{
int sum = 0;
int x = number;
while (x != 0) {
int last = x % 10;
sum += factorial (last);
x /= 10;
}
if (sum == number)
return "STRONG!!!!";
else
return "Not Strong !!";
}
I tried this Codility test: MinAbsSum.
https://codility.com/programmers/lessons/17-dynamic_programming/min_abs_sum/
I solved the problem by searching the whole tree of possibilities. The results were OK, however, my solution failed due to timeout for large input. In other words the time complexity was not as good as expected. My solution is O(nlogn), something normal with trees. But this coding test was in the section "Dynamic Programming", and there must be some way to improve it. I tried with summing the whole set first and then using this information, but always there is something missing in my solution. Does anybody have an idea on how to improve my solution using DP?
#include <vector>
using namespace std;
int sum(vector<int>& A, size_t i, int s)
{
if (i == A.size())
return s;
int tmpl = s + A[i];
int tmpr = s - A[i];
return min (abs(sum(A, i+1, tmpl)), abs(sum(A, i+1, tmpr)));
}
int solution(vector<int> &A) {
return sum(A, 0, 0);
}
I could not solve it. But here's the official answer.
Quoting it:
Notice that the range of numbers is quite small (maximum 100). Hence,
there must be a lot of duplicated numbers. Let count[i] denote the
number of occurrences of the value i. We can process all occurrences
of the same value at once. First we calculate values count[i] Then we
create array dp such that:
dp[j] = −1 if we cannot get the sum j,
dp[j] >= 0 if we can get sum j.
Initially, dp[j] = -1 for all of j (except dp[0] = 0). Then we scan
through all the values a appearing in A; we consider all a such
that count[a]>0. For every such a we update dp that dp[j] denotes
how many values a remain (maximally) after achieving sum j. Note
that if the previous value at dp[j] >= 0 then we can set dp[j] =
count[a] as no value a is needed to obtain the sum j. Otherwise we
must obtain sum j-a first and then use a number a to get sum j. In
such a situation dp[j] = dp[j-a]-1. Using this algorithm, we can
mark all the sum values and choose the best one (closest to half of S,
the sum of abs of A).
def MinAbsSum(A):
N = len(A)
M = 0
for i in range(N):
A[i] = abs(A[i])
M = max(A[i], M)
S = sum(A)
count = [0] * (M + 1)
for i in range(N):
count[A[i]] += 1
dp = [-1] * (S + 1)
dp[0] = 0
for a in range(1, M + 1):
if count[a] > 0:
for j in range(S):
if dp[j] >= 0:
dp[j] = count[a]
elif (j >= a and dp[j - a] > 0):
dp[j] = dp[j - a] - 1
result = S
for i in range(S // 2 + 1):
if dp[i] >= 0:
result = min(result, S - 2 * i)
return result
(note that since the final iteration only considers sums up until S // 2 + 1, we can save some space and time by only creating a DP Cache up until that value as well)
The Java answer provided by fladam returns wrong result for input [2, 3, 2, 2, 3], although it gets 100% score.
Java Solution
import java.util.Arrays;
public class MinAbsSum{
static int[] dp;
public static void main(String args[]) {
int[] array = {1, 5, 2, -2};
System.out.println(findMinAbsSum(array));
}
public static int findMinAbsSum(int[] A) {
int arrayLength = A.length;
int M = 0;
for (int i = 0; i < arrayLength; i++) {
A[i] = Math.abs(A[i]);
M = Math.max(A[i], M);
}
int S = sum(A);
dp = new int[S + 1];
int[] count = new int[M + 1];
for (int i = 0; i < arrayLength; i++) {
count[A[i]] += 1;
}
Arrays.fill(dp, -1);
dp[0] = 0;
for (int i = 1; i < M + 1; i++) {
if (count[i] > 0) {
for(int j = 0; j < S; j++) {
if (dp[j] >= 0) {
dp[j] = count[i];
} else if (j >= i && dp[j - i] > 0) {
dp[j] = dp[j - i] - 1;
}
}
}
}
int result = S;
for (int i = 0; i < Math.floor(S / 2) + 1; i++) {
if (dp[i] >= 0) {
result = Math.min(result, S - 2 * i);
}
}
return result;
}
public static int sum(int[] array) {
int sum = 0;
for(int i : array) {
sum += i;
}
return sum;
}
}
I invented another solution, better than the previous one. I do not use recursion any more.
This solution works OK (all logical tests passed), and also passed some of the performance tests, but not all. How else can I improve it?
#include <vector>
#include <set>
using namespace std;
int solution(vector<int> &A) {
if (A.size() == 0) return 0;
set<int> sums, tmpSums;
sums.insert(abs(A[0]));
for (auto it = begin(A) + 1; it != end(A); ++it)
{
for (auto s : sums)
{
tmpSums.insert(abs(s + abs(*it)));
tmpSums.insert(abs(s - abs(*it)));
}
sums = tmpSums;
tmpSums.clear();
}
return *sums.begin();
}
This solution (in Java) scored 100% for both (correctness and performance)
public int solution(int[] a){
if (a.length == 0) return 0;
if (a.length == 1) return a[0];
int sum = 0;
for (int i=0;i<a.length;i++){
sum += Math.abs(a[i]);
}
int[] indices = new int[a.length];
indices[0] = 0;
int half = sum/2;
int localSum = Math.abs(a[0]);
int minLocalSum = Integer.MAX_VALUE;
int placeIndex = 1;
for (int i=1;i<a.length;i++){
if (localSum<half){
if (Math.abs(2*minLocalSum-sum) > Math.abs(2*localSum - sum))
minLocalSum = localSum;
localSum += Math.abs(a[i]);
indices[placeIndex++] = i;
}else{
if (localSum == half)
return Math.abs(2*half - sum);
if (Math.abs(2*minLocalSum-sum) > Math.abs(2*localSum - sum))
minLocalSum = localSum;
if (placeIndex > 1) {
localSum -= Math.abs(a[indices[placeIndex--]]);
i = indices[placeIndex];
}
}
}
return (Math.abs(2*minLocalSum - sum));
}
this solution treats all elements like they are positive numbers and it's looking to reach as close as it can to the sum of all elements divided by 2 (in that case we know that the sum of all other elements will be the same delta far from the half too -> abs sum will be minimum possible ).
it does so by starting with the first element and successively adding others to the "local" sum (and recording indices of elements in the sum) until it reaches sum of x >= sumAll/2. if that x is equal to sumAll/2 we have an optimal solution. if not, we go step back in the indices array and continue picking other element where last iteration in that position ended. the result will be a "local" sum having abs((sumAll - sum) - sum) closest to 0;
fixed solution:
public static int solution(int[] a){
if (a.length == 0) return 0;
if (a.length == 1) return a[0];
int sum = 0;
for (int i=0;i<a.length;i++) {
a[i] = Math.abs(a[i]);
sum += a[i];
}
Arrays.sort(a);
int[] arr = a;
int[] arrRev = new int[arr.length];
int minRes = Integer.MAX_VALUE;
for (int t=0;t<=4;t++) {
arr = fold(arr);
int res1 = findSum(arr, sum);
if (res1 < minRes) minRes = res1;
rev(arr, arrRev);
int res2 = findSum(arrRev, sum);
if (res2 < minRes) minRes = res2;
arrRev = fold(arrRev);
int res3 = findSum(arrRev, sum);
if (res3 < minRes) minRes = res3;
}
return minRes;
}
private static void rev(int[] arr, int[] arrRev){
for (int i = 0; i < arrRev.length; i++) {
arrRev[i] = arr[arr.length - 1 - i];
}
}
private static int[] fold(int[] a){
int[] arr = new int[a.length];
for (int i=0;a.length/2+i/2 < a.length && a.length/2-i/2-1 >= 0;i+=2){
arr[i] = a[a.length/2+i/2];
arr[i+1] = a[a.length/2-i/2-1];
}
if (a.length % 2 > 0) arr[a.length-1] = a[a.length-1];
else{
arr[a.length-2] = a[0];
arr[a.length-1] = a[a.length-1];
}
return arr;
}
private static int findSum(int[] arr, int sum){
int[] indices = new int[arr.length];
indices[0] = 0;
double half = Double.valueOf(sum)/2;
int localSum = Math.abs(arr[0]);
int minLocalSum = Integer.MAX_VALUE;
int placeIndex = 1;
for (int i=1;i<arr.length;i++){
if (localSum == half)
return 2*localSum - sum;
if (Math.abs(2*minLocalSum-sum) > Math.abs(2*localSum - sum))
minLocalSum = localSum;
if (localSum<half){
localSum += Math.abs(arr[i]);
indices[placeIndex++] = i;
}else{
if (placeIndex > 1) {
localSum -= Math.abs(arr[indices[--placeIndex]]);
i = indices[placeIndex];
}
}
}
return Math.abs(2*minLocalSum - sum);
}
The following is a rendering of the official answer in C++ (scoring 100% in task, correctness, and performance):
#include <cmath>
#include <algorithm>
#include <numeric>
using namespace std;
int solution(vector<int> &A) {
// write your code in C++14 (g++ 6.2.0)
const int N = A.size();
int M = 0;
for (int i=0; i<N; i++) {
A[i] = abs(A[i]);
M = max(M, A[i]);
}
int S = accumulate(A.begin(), A.end(), 0);
vector<int> counts(M+1, 0);
for (int i=0; i<N; i++) {
counts[A[i]]++;
}
vector<int> dp(S+1, -1);
dp[0] = 0;
for (int a=1; a<M+1; a++) {
if (counts[a] > 0) {
for (int j=0; j<S; j++) {
if (dp[j] >= 0) {
dp[j] = counts[a];
} else if ((j >= a) && (dp[j-a] > 0)) {
dp[j] = dp[j-a]-1;
}
}
}
}
int result = S;
for (int i =0; i<(S/2+1); i++) {
if (dp[i] >= 0) {
result = min(result, S-2*i);
}
}
return result;
}
You are almost 90% to the actual solution. It seems you understand recursion very well. Now, You should apply dynamic programming here with your program.
Dynamic Programming is nothing but memoization to the recursion so that we will not calculate same sub problems again and again. If same sub problems encounter , we return the previously calculated and memorized value. Memorization can be done with the help of a 2D array , say dp[][], where first state represent current index of array and second state represent summation.
For this problem specific, instead of giving calls to both states from each state, you sometimes can greedily take decision to skip one call.
I would like to provide the algorithm and then my implementation in C++. Idea is more or less the same as the official codility solution with some constant optimisation added.
Calculate the maximum absolute element of the inputs.
Calculate the absolute sum of the inputs.
Count the number of occurrence of each number in the inputs. Store the results in a vector hash.
Go through each input.
For each input, goes through all possible sums of any number of inputs. It is a slight constant optimisation to go only up to half of the possible sums.
For each sum that has been made before, set the occurrence count of the current input.
Check for each potential sum equal to or greater than the current input whether this input has already been used before. Update the values at the current sum accordingly. We do not need to check for potential sums less than the current input in this iteration, since it is evident that it has not been used before.
The above nested loop will fill in each possible sum with a value greater than -1.
Go through this possible sum hash again to look for the closest sum to half that is possible to make. Eventually, the min abs sum will be the difference of this from the half multiplied by two as the difference will be added up in both groups as the difference from the median.
The runtime complexity of this algorithm is O(N * max(abs(A)) ^ 2), or simply O(N * M ^ 2). That is because the outer loop is iterating M times and the inner loop is iterating sum times. The sum is basically N * M in worst case. Therefore, it is O(M * N * M).
The space complexity of this solution is O(N * M) because we allocate a hash of N items for the counts and a hash of S items for the sums. S is N * M again.
int solution(vector<int> &A)
{
int M = 0, S = 0;
for (const int e : A) { M = max(abs(e), M); S += abs(e); }
vector<int> counts(M + 1, 0);
for (const int e : A) { ++counts[abs(e)]; }
vector<int> sums(S + 1, -1);
sums[0] = 0;
for (int ci = 1; ci < counts.size(); ++ci) {
if (!counts[ci]) continue;
for (int si = 0; si < S / 2 + 1; ++si) {
if (sums[si] >= 0) sums[si] = counts[ci];
else if (si >= ci and sums[si - ci] > 0) sums[si] = sums[si - ci] - 1;
}
}
int min_abs_sum = S;
for (int i = S / 2; i >= 0; --i) if (sums[i] >= 0) return S - 2 * i;
return min_abs_sum;
}
Let me add my 50 cent, how to come up with the score 100% solution.
For me it was hard to understand the ultimate solution, proposed earlier in this thread.
So I started with warm-up solution with score 63%, because its O(NxNxM),
and because it doesn't use the fact that M is quite small value, and there are many duplicates in big arrays
here the key part is to understand how array isSumPossible is filled and interpreted:
how to fill array isSumPossible using numbers in input array:
if isSumPossible[sum] >= 0, i.e. sum is already possible, even without current number, then let's set it's value to 1 - count of current number, that is left unused for this sum, it'll go to our "reserve", so we can use it later for greater sums.
if (isSumPossible[sum] >= 0) {
isSumPossible[sum] = 1;
}
if isSumPossible[sum] <= 0, i.e. sum is considered not yet possible, with all input numbers considered previously, then let's check maybe
smaller sum sum - number is already considered as possible, and we have in "reserve" our current number (isSumPossible[sum - number] == 1), then do following
else if (sum >= number && isSumPossible[sum - number] == 1) {
isSumPossible[sum] = 0;
}
here isSumPossible[sum] = 0 means that we have used number in composing sum and it's now considered as possible (>=0), but we have no number in "reserve", because we've used it ( =0)
how to interpret filled array isSumPossible after considering all numbers in input array:
if isSumPossible[sum] >= 0 then the sum is possible, i.e. it can be reached by summation of some numbers in given array
if isSumPossible[sum] < 0 then the sum can't be reached by summation of any numbers in given array
The more simple thing here is to understand why we are searching sums only in interval [0, maxSum/2]:
because if find a possible sum, that is very close to maxSum/2,
ideal case here if we've found possible sum = maxSum/2,
if so, then it's obvious, that we can somehow use the rest numbers in input array to make another maxSum/2, but now with negative sign, so as a result of annihilation we'll get solution = 0, because maxSum/2 + (-1)maxSum/2 = 0.
But 0 the best case solution, not always reachable.
But we, nevertheless, should seek for the minimal delta = ((maxSum - sum) - sum),
so this we seek for delta -> 0, that's why we have this:
int result = Integer.MAX_VALUE;
for (int sum = 0; sum < maxSum / 2 + 1; sum++) {
if (isSumPossible[sum] >= 0) {
result = Math.min(result, (maxSum - sum) - sum);
}
}
warm-up solution
public int solution(int[] A) {
if (A == null || A.length == 0) {
return 0;
}
if (A.length == 1) {
return A[0];
}
int maxSum = 0;
for (int i = 0; i < A.length; i++) {
A[i] = Math.abs(A[i]);
maxSum += A[i];
}
int[] isSumPossible = new int[maxSum + 1];
Arrays.fill(isSumPossible, -1);
isSumPossible[0] = 0;
for (int number : A) {
for (int sum = 0; sum < maxSum / 2 + 1; sum++) {
if (isSumPossible[sum] >= 0) {
isSumPossible[sum] = 1;
} else if (sum >= number && isSumPossible[sum - number] == 1) {
isSumPossible[sum] = 0;
}
}
}
int result = Integer.MAX_VALUE;
for (int sum = 0; sum < maxSum / 2 + 1; sum++) {
if (isSumPossible[sum] >= 0) {
result = Math.min(result, maxSum - 2 * sum);
}
}
return result;
}
and after this we can optimize it, using the fact that there are many duplicate numbers in big arrays, and we come up with the solution with 100% score, its O(Mx(NxM)), because maxSum = NxM at worst case
public int solution(int[] A) {
if (A == null || A.length == 0) {
return 0;
}
if (A.length == 1) {
return A[0];
}
int maxNumber = 0;
int maxSum = 0;
for (int i = 0; i < A.length; i++) {
A[i] = Math.abs(A[i]);
maxNumber = Math.max(maxNumber, A[i]);
maxSum += A[i];
}
int[] count = new int[maxNumber + 1];
for (int i = 0; i < A.length; i++) {
count[A[i]]++;
}
int[] isSumPossible = new int[maxSum + 1];
Arrays.fill(isSumPossible, -1);
isSumPossible[0] = 0;
for (int number = 0; number < maxNumber + 1; number++) {
if (count[number] > 0) {
for (int sum = 0; sum < maxSum / 2 + 1; sum++) {
if (isSumPossible[sum] >= 0) {
isSumPossible[sum] = count[number];
} else if (sum >= number && isSumPossible[sum - number] > 0) {
isSumPossible[sum] = isSumPossible[sum - number] - 1;
}
}
}
}
int result = Integer.MAX_VALUE;
for (int sum = 0; sum < maxSum / 2 + 1; sum++) {
if (isSumPossible[sum] >= 0) {
result = Math.min(result, maxSum - 2 * sum);
}
}
return result;
}
I hope I've made it at least a little clear
Kotlin solution
Time complexity: O(N * max(abs(A))**2)
Score: 100%
import kotlin.math.*
fun solution(A: IntArray): Int {
val N = A.size
var M = 0
for (i in 0 until N) {
A[i] = abs(A[i])
M = max(M, A[i])
}
val S = A.sum()
val counts = MutableList(M + 1) { 0 }
for (i in 0 until N) {
counts[A[i]]++
}
val dp = MutableList(S + 1) { -1 }
dp[0] = 0
for (a in 1 until M + 1) {
if (counts[a] > 0) {
for (j in 0 until S) {
if (dp[j] >= 0) {
dp[j] = counts[a]
} else if (j >= a && dp[j - a] > 0) {
dp[j] = dp[j - a] - 1
}
}
}
}
var result = S
for (i in 0 until (S / 2 + 1)) {
if (dp[i] >= 0) {
result = minOf(result, S - 2 * i)
}
}
return result
}
I've tried to check whether a number is a palindrome with the following code:
unsigned short digitsof (unsigned int x)
{
unsigned short n = 0;
while (x)
{
x /= 10;
n++;
}
return n;
}
bool ispalindrome (unsigned int x)
{
unsigned short digits = digitsof (x);
for (unsigned short i = 1; i <= digits / 2; i++)
{
if (x % (unsigned int)pow (10, i) != x % (unsigned int)pow (10, digits - 1 + i))
{
return false;
}
}
return true;
}
However, the following code isn't able to check for palindromes - false is always returned even if the number is a palindrome.
Can anyone point out the error?
(Please note: I'm not interested to make it into a string and reverse it to see where the problem is: rather, I'm interested to know where the error is in the above code.)
I personally would just build a string from the number, and then treat it as a normal palindrome check (check that each character in the first half matches the ones at length()-index).
x % (unsigned int)pow (10, i) is not the ith digit.
The problem is this:
x % (unsigned int)pow (10, i)
Lets try:
x =504405
i =3
SO I want 4.
x % 10^3 => 504405 %1000 => 405 NOT 4
How about
x / (unsigned int)pow (10, i -1) % 10
Just for more info! The following two functions are working for me:
double digitsof (double x)
{
double n = 0;
while (x > 1)
{
x /= 10;
n++;
}
return n;
}
bool ispalindrome (double x)
{
double digits = digitsof (x);
double temp = x;
for(double i = 1; i <= digits/2; i++)
{
float y = (int)temp % 10;
cout<<y<<endl;
temp = temp/10;
float z = (int)x / (int)pow(10 , digits - i);
cout<<(int)z<<endl;
x = (int)x % (int)pow(10 , digits - i);
if(y != z)
return false;
}
return true;
}
Code to check if given number is palindrome or not in JAVA
import java.util.*;
public class HelloWorld{
private static int countDigits(int num) {
int count = 0;
while(num>0) {
count++;
num /= 10;
}
return count;
}
public static boolean isPalin(int num) {
int digs = HelloWorld.countDigits(num);
int divderToFindMSD = 1;
int divderToFindLSD = 1;
for (int i = 0; i< digs -1; i++)
divderToFindMSD *= 10;
int mid = digs/2;
while(mid-- != 0)
{
int msd = (num/divderToFindMSD)%10;
int lsd = (num/divderToFindLSD)%10;
if(msd!=lsd)
return false;
divderToFindMSD /= 10;
divderToFindLSD *= 10;
}
return true;
}
public static void main(String []args) {
boolean isPalin = HelloWorld.isPalin(1221);
System.out.println("Results: " + isPalin);
}
}
I have done this with my own solution which is restricted with these conditions
Do not convert int to string.
Do not use any helper function.
var inputNumber = 10801
var firstDigit = 0
var lastDigit = 0
var quotient = inputNumber
while inputNumber > 0 {
lastDigit = inputNumber % 10
var tempNum = inputNumber
var count = 0
while tempNum > 0 {
tempNum = tempNum / 10
count = count + 1
}
var n = 1
for _ in 1 ..< count {
n = n * 10
}
firstDigit = quotient / n
if firstDigit != lastDigit {
print("Not a palindrome :( ")
break
}
quotient = quotient % n
inputNumber = inputNumber / 10
}
if firstDigit == lastDigit {
print("It's a palindrome :D :D ")
}
Im trying to implement the Miller-Rabin primality test according to the description in FIPS 186-3 C.3.1. No matter what I do, I cannot get it to work. The instructions are pretty specific, and I dont think I missed anything, and yet Im getting true for non-prime values.
What did I do wrong?
template <typename R, typename S, typename T>
T POW(R base, S exponent, const T mod){
T result = 1;
while (exponent){
if (exponent & 1)
result = (result * base) % mod;
exponent >>= 1;
base = (base * base) % mod;
}
return result;
}
// used uint64_t to prevent overflow, but only testing with small numbers for now
bool MillerRabin_FIPS186(uint64_t w, unsigned int iterations = 50){
srand(time(0));
unsigned int a = 0;
uint64_t W = w - 1; // dont want to keep calculating w - 1
uint64_t m = W;
while (!(m & 1)){
m >>= 1;
a++;
}
// skipped getting wlen
// when i had this function using my custom arbitrary precision integer class,
// and could get len(w), getting it and using it in an actual RBG
// made no difference
for(unsigned int i = 0; i < iterations; i++){
uint64_t b = (rand() % (W - 3)) + 2; // 2 <= b <= w - 2
uint64_t z = POW(b, m, w);
if ((z == 1) || (z == W))
continue;
else
for(unsigned int j = 1; j < a; j++){
z = POW(z, 2, w);
if (z == W)
continue;
if (z == 1)
return 0;// Composite
}
}
return 1;// Probably Prime
}
this:
std::cout << MillerRabin_FIPS186(33) << std::endl;
std::cout << MillerRabin_FIPS186(35) << std::endl;
std::cout << MillerRabin_FIPS186(37) << std::endl;
std::cout << MillerRabin_FIPS186(39) << std::endl;
std::cout << MillerRabin_FIPS186(45) << std::endl;
std::cout << MillerRabin_FIPS186(49) << std::endl;
is giving me:
0
1
1
1
0
1
The only difference between your implementation and Wikipedia's is that you forgot the second return composite statement. You should have a return 0 at the end of the loop.
Edit: As pointed out by Daniel, there is a second difference. The continue is continuing the inner loop, rather than the outer loop like it's supposed to.
for(unsigned int i = 0; i < iterations; i++){
uint64_t b = (rand() % (W - 3)) + 2; // 2 <= b <= w - 2
uint64_t z = POW(b, m, w);
if ((z == 1) || (z == W))
continue;
else{
int continueOuter = 0;
for(unsigned int j = 1; j < a; j++){
z = POW(z, 2, w);
if (z == W)
continueOuter = 1;
break;
if (z == 1)
return 0;// Composite
}
if (continueOuter) {continue;}
}
return 0; //This is the line you're missing.
}
return 1;// Probably Prime
Also, if the input is even, it will always return probably prime since a is 0. You should add an extra check at the start for that.
In the inner loop,
for(unsigned int j = 1; j < a; j++){
z = POW(z, 2, w);
if (z == W)
continue;
if (z == 1)
return 0;// Composite
}
you should break; instead of continue; when z == W. By continueing, in the next iteration of that loop, if there is one, z will become 1 and the candidate is possibly wrongly declared composite. Here, that happens for 17, 41, 73, 89 and 97 among the primes less than 100.
I'm working on Project Euler #27 in C++:
Euler published the remarkable quadratic formula:
n² + n + 41
It turns out that the formula will produce 40 primes for the
consecutive values n = 0 to 39. However, when n = 40, 40² + 40 + 41 =
40(40 + 1) + 41 is divisible by 41, and certainly when n = 41, 41² +
41 + 41 is clearly divisible by 41.
Using computers, the incredible formula n² − 79n + 1601 was
discovered, which produces 80 primes for the consecutive values n = 0
to 79. The product of the coefficients, −79 and 1601, is −126479.
Considering quadratics of the form:
n² + an + b, where |a| < 1000 and |b| < 1000
where |n| is the modulus/absolute value of n
e.g. |11| = 11 and |−4| = 4
Find the product of the coefficients, a and b, for the quadratic
expression that produces the maximum number of primes for consecutive
values of n, starting with n = 0.
I keep getting -60939 when the real answer is -59231. What am I missing?
#include <iostream>
#include "Helper.h"
using namespace std;
int formula(int a, int b, int n) {
return ((n * n) + (a * n) + b);
}
int main() {
int most = 0;
int ansA = 0;
int ansB = 0;
bool end = false;
for(int a = 999; a >= -999; a--) {
for(int b = 999; b >= 2; b--) { //b must be prime
if(Helper::isPrime(b)) {
end = false;
for(int n = 0; !end; n++) {
if(!Helper::isPrime(formula(a, b, n))) {
if(n-1 > most) {
most = n-1;
ansA = a;
ansB = b;
}
end = true;
}
}
}
}
}
cout << ansA << " * " << ansB << " = " << ansA * ansB << " with " << most << " primes." << endl;
return 0;
}
In case it's the problem, here is my isPrime function:
bool Helper::isPrime(int num) {
if(num == 2)
return true;
if(num % 2 == 0 || num == 1 || num == 0)
return false;
int root = (int) sqrt((double)num) + 1;
for(int i = root; i >= 2; i--) {
if (num % i == 0)
return false;
}
return true;
}
You are allowing a to be negative, and your formula returns an int. Does calling Helper::isPrime with a negative number even make sense (in other words, does Helper::isPrime take an unsigned int?)
Here is my java version. Hope it helps:
static int function(int n, int a, int b){
return n*n + a*n + b;
}
static int consequitive_Primes(int a, int b, HashSet<Integer> primes){
int n = 0;
int number = 0;
while(true){
if(!primes.contains(function(n, a, b)))
break;
number++;
n++;
}
return number;
}
static HashSet<Integer> primes (int n){
ArrayList<Integer> primes = new ArrayList<Integer>();
primes.add(3);
for(int i=3; i<n;i+=2){
boolean isPrime = true;
for(Integer k:primes){
if(i%k==0){
isPrime = false;
break;
}
}
if(isPrime) primes.add(i);
}
return new HashSet<Integer>(primes);
}
static long q27(){
HashSet<Integer> primes = primes(1000);
int max = 0;
int max_ab = 0;
for(int a = -999; a<1000;a++){
for(int b = -999; b<1000;b++){
int prime_No = consequitive_Primes(a,b,primes);
if(max<prime_No){
max = prime_No;
max_ab = a*b;
}
}
}
return max_ab;
}