I want to find the sum of all the positive integers in the range [1, N] with a given digit sum d. For example, if n = 100 and d = 7, the answer will be 7 + 16 + 25 + 34 + 43 + 52 + 61 + 70 = 308.
Following code can be used to count the numbers in the range [1, N] with a given digit sum d.
cnt[i][0][s] denotes count of suffixes that can be formed starting from index i, whose digits add up to s.
cnt[i][1][s] count of suffixes that can be formed starting from index i, whose digits add up to s such that the formed suffix is not greater than corresponding suffix in input string
#include <bits/stdc++.h>
using namespace std;
typedef long long int i64;
i64 cnt[20][2][200];
void digit_sum_dp(string ss) {
int n = ss.size();
for (int i = 0; i < 20; i++) {
for (int j = 0; j < 2; j++) {
for (int k = 0; k < 200; k++) {
cnt[i][j][k] = 0;
}
}
}
cnt[n][0][0] = 1;
cnt[n][1][0] = 1;
for (int i = n - 1; i >= 0; i--) {
for (int tight = 0; tight < 2; tight++) {
for (int sum = 0; sum < 200; sum++) {
if (tight) {
for (int d = 0; d <= ss[i] - '0'; d++) {
if (d == ss[i] - '0') {
cnt[i][1][sum] += cnt[i + 1][1][sum - d];
} else {
cnt[i][1][sum] += cnt[i + 1][0][sum - d];
}
}
} else {
for (int d = 0; d < 10; d++) {
cnt[i][0][sum] += cnt[i + 1][0][sum - d];
}
}
}
}
}
return cnt[0][1][d];
}
int main() {
string str = "100";
int d = 7;
cout << digit_sum_dp(str, d) << "\n";
return 0;
}
I have tried to extend the code to find out the sum of numbers instead of the count of numbers. Following is a code snippet.
cnt[i][1][sum] += cnt[i + 1][1][sum - d];
tot[i][1][sum] += (d * cnt[i + 1][1][sum - d] + tot[i + 1][1][sum - d] * pow(10, i));
I am getting incorrect results for some of the inputs. I shall be grateful if someone can help me.
I am attempting to make a multithreaded program that computes the prime-factor of each input(each input having its own thread). Using realloc or calloc with a int* Name, inputing large numbers or lots of thread {1..100} gives me realloc:invalid size or malloc.c failure. I am not sure if it is a deep pointer issue, but I need an output of such:
./thread {1..10}
1:
2: 2
3: 3
4: 2 2
5: 5
etc....
10: 2 5
int size = 5;
int* primeNumbers = (int*)calloc(size,sizeof(int));
int n = atoi(param);
//sum = 0;
int counter = 0;
//Add original value to int* primeNumbers for return printing
primeNumbers[counter] = n;
counter++;
while (n % 2 == 0)
{
printf("%d ", 2);
//If max allocated memory is reached, double memory size
if (counter == size)
{
size = size * 2;
primeNumbers = realloc(primeNumbers,size*sizeof(int));
primeNumbers[size] = 0;
}
primeNumbers[counter] = 2;
counter++;
n = n/2;
}
for (int i = 3; i <= sqrt(n); i = i + 2)
{
while (n % i == 0)
{
printf("%d ",i);
//If max allocated memory is reached, double memory size
if (counter == size)
{
size = size * 2;
primeNumbers = realloc(primeNumbers,size*sizeof(int));
primeNumbers[size] = 0;
}
primeNumbers[counter] = i;
counter++;
n = n/i;
}
}
enter code here
I have the following errors
:( identifies 5105105105105100 as MASTERCARD
expected "MASTERCARD\n", not "INVALID\n"
:( identifies 4111111111111111 as VISA
expected "VISA\n", not "INVALID\n"
:( identifies 4012888888881881 as VISA
expected "VISA\n", not "INVALID\n"
but for the card to be correct the last digit should be which is correct in my case. Please help
------------------- CODE -----------------------
#include <cs50.h>
#include <stdio.h>
int main()
{
long long cardNumber;
// get a card number from the user
do
{
printf("Your card number please: ");
//scanf("%lld", &cardNumber);
cardNumber = get_long_long();
}
while (cardNumber < 0);
//check the length of the card
int counter = 0;
long long cardNumberNeo = cardNumber;
while (cardNumberNeo > 0)
{
cardNumberNeo = cardNumberNeo / 10;
counter++;
}
if (counter != 15 && counter != 16 && counter != 13)
{
printf("INVALID\n");
}
// Array of card number
cardNumberNeo = cardNumber;
int cardNumberArr[counter], cardNumberArrNeo[counter], i;
for (i=0; i<counter; i++)
{
cardNumberArr[counter-i-1] = cardNumberNeo % 10;
cardNumberArrNeo[counter-i-1] = cardNumberArr[counter-i-1];
cardNumberNeo = cardNumberNeo / 10;
}
for (int i = 1; i < counter; i+=2)
{
cardNumberArrNeo[i] = cardNumberArrNeo[i] * 2;
}
int oddNumber = 0;
int temp;
for (int i = 0; i < counter; i++)
{
temp = (cardNumberArrNeo[i] % 10) + (cardNumberArrNeo[i]/10 % 10);
oddNumber = oddNumber + temp;
}
if (oddNumber % 10 == 0)
{
// Check the type of the card
if ( ((cardNumberArr[0] == 3 && cardNumberArr[1] == 4) || (cardNumberArr[0] == 3 && cardNumberArr[1] == 7)) && counter == 15 )
{
printf("AMEX\n");
}
else if (cardNumberArr[0] == 5 && cardNumberArr[1] >= 1 && cardNumberArr[1] <= 5 && counter == 16)
{
printf("MASTERCARD\n");
}
else if (cardNumberArr[0] == 4 && (counter == 13 || counter == 16 ))
{
printf("VISA\n");
}
else
{
printf("INVALID\n");
}
}
else
{
printf("INVALID\n");
}
return 0;
}
This will work.....as long as the credit card number has an odd length. Remember, Luhn's algorithm rule [emphasis added]:
Multiply every other digit by 2, starting with the number’s
second-to-last digit, and then add those products' digits together.
This loop for (int i = 1; i < counter; i+=2) processes the wrong digits for a card with even length. Such numbers need to start at the 0th index in order to end up at the correct place (ie penultimate digit).
I'm posting this although much has already been posted about this question. I didn't want to post as an answer since it's not working. The answer to this post (Finding the rank of the Given string in list of all possible permutations with Duplicates) did not work for me.
So I tried this (which is a compilation of code I've plagiarized and my attempt to deal with repetitions). The non-repeating cases work fine. BOOKKEEPER generates 83863, not the desired 10743.
(The factorial function and letter counter array 'repeats' are working correctly. I didn't post to save space.)
while (pointer != length)
{
if (sortedWordChars[pointer] != wordArray[pointer])
{
// Swap the current character with the one after that
char temp = sortedWordChars[pointer];
sortedWordChars[pointer] = sortedWordChars[next];
sortedWordChars[next] = temp;
next++;
//For each position check how many characters left have duplicates,
//and use the logic that if you need to permute n things and if 'a' things
//are similar the number of permutations is n!/a!
int ct = repeats[(sortedWordChars[pointer]-64)];
// Increment the rank
if (ct>1) { //repeats?
System.out.println("repeating " + (sortedWordChars[pointer]-64));
//In case of repetition of any character use: (n-1)!/(times)!
//e.g. if there is 1 character which is repeating twice,
//x* (n-1)!/2!
int dividend = getFactorialIter(length - pointer - 1);
int divisor = getFactorialIter(ct);
int quo = dividend/divisor;
rank += quo;
} else {
rank += getFactorialIter(length - pointer - 1);
}
} else
{
pointer++;
next = pointer + 1;
}
}
Note: this answer is for 1-based rankings, as specified implicitly by example. Here's some Python that works at least for the two examples provided. The key fact is that suffixperms * ctr[y] // ctr[x] is the number of permutations whose first letter is y of the length-(i + 1) suffix of perm.
from collections import Counter
def rankperm(perm):
rank = 1
suffixperms = 1
ctr = Counter()
for i in range(len(perm)):
x = perm[((len(perm) - 1) - i)]
ctr[x] += 1
for y in ctr:
if (y < x):
rank += ((suffixperms * ctr[y]) // ctr[x])
suffixperms = ((suffixperms * (i + 1)) // ctr[x])
return rank
print(rankperm('QUESTION'))
print(rankperm('BOOKKEEPER'))
Java version:
public static long rankPerm(String perm) {
long rank = 1;
long suffixPermCount = 1;
java.util.Map<Character, Integer> charCounts =
new java.util.HashMap<Character, Integer>();
for (int i = perm.length() - 1; i > -1; i--) {
char x = perm.charAt(i);
int xCount = charCounts.containsKey(x) ? charCounts.get(x) + 1 : 1;
charCounts.put(x, xCount);
for (java.util.Map.Entry<Character, Integer> e : charCounts.entrySet()) {
if (e.getKey() < x) {
rank += suffixPermCount * e.getValue() / xCount;
}
}
suffixPermCount *= perm.length() - i;
suffixPermCount /= xCount;
}
return rank;
}
Unranking permutations:
from collections import Counter
def unrankperm(letters, rank):
ctr = Counter()
permcount = 1
for i in range(len(letters)):
x = letters[i]
ctr[x] += 1
permcount = (permcount * (i + 1)) // ctr[x]
# ctr is the histogram of letters
# permcount is the number of distinct perms of letters
perm = []
for i in range(len(letters)):
for x in sorted(ctr.keys()):
# suffixcount is the number of distinct perms that begin with x
suffixcount = permcount * ctr[x] // (len(letters) - i)
if rank <= suffixcount:
perm.append(x)
permcount = suffixcount
ctr[x] -= 1
if ctr[x] == 0:
del ctr[x]
break
rank -= suffixcount
return ''.join(perm)
If we use mathematics, the complexity will come down and will be able to find rank quicker. This will be particularly helpful for large strings.
(more details can be found here)
Suggest to programmatically define the approach shown here (screenshot attached below) given below)
I would say David post (the accepted answer) is super cool. However, I would like to improve it further for speed. The inner loop is trying to find inverse order pairs, and for each such inverse order, it tries to contribute to the increment of rank. If we use an ordered map structure (binary search tree or BST) in that place, we can simply do an inorder traversal from the first node (left-bottom) until it reaches the current character in the BST, rather than traversal for the whole map(BST). In C++, std::map is a perfect one for BST implementation. The following code reduces the necessary iterations in loop and removes the if check.
long long rankofword(string s)
{
long long rank = 1;
long long suffixPermCount = 1;
map<char, int> m;
int size = s.size();
for (int i = size - 1; i > -1; i--)
{
char x = s[i];
m[x]++;
for (auto it = m.begin(); it != m.find(x); it++)
rank += suffixPermCount * it->second / m[x];
suffixPermCount *= (size - i);
suffixPermCount /= m[x];
}
return rank;
}
#Dvaid Einstat, this was really helpful. It took me a WHILE to figure out what you were doing as I am still learning my first language(C#). I translated it into C# and figured that I'd give that solution as well since this listing helped me so much!
Thanks!
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
using System.Text.RegularExpressions;
namespace CsharpVersion
{
class Program
{
//Takes in the word and checks to make sure that the word
//is between 1 and 25 charaters inclusive and only
//letters are used
static string readWord(string prompt, int high)
{
Regex rgx = new Regex("^[a-zA-Z]+$");
string word;
string result;
do
{
Console.WriteLine(prompt);
word = Console.ReadLine();
} while (word == "" | word.Length > high | rgx.IsMatch(word) == false);
result = word.ToUpper();
return result;
}
//Creates a sorted dictionary containing distinct letters
//initialized with 0 frequency
static SortedDictionary<char,int> Counter(string word)
{
char[] wordArray = word.ToCharArray();
int len = word.Length;
SortedDictionary<char,int> count = new SortedDictionary<char,int>();
foreach(char c in word)
{
if(count.ContainsKey(c))
{
}
else
{
count.Add(c, 0);
}
}
return count;
}
//Creates a factorial function
static int Factorial(int n)
{
if (n <= 1)
{
return 1;
}
else
{
return n * Factorial(n - 1);
}
}
//Ranks the word input if there are no repeated charaters
//in the word
static Int64 rankWord(char[] wordArray)
{
int n = wordArray.Length;
Int64 rank = 1;
//loops through the array of letters
for (int i = 0; i < n-1; i++)
{
int x=0;
//loops all letters after i and compares them for factorial calculation
for (int j = i+1; j<n ; j++)
{
if (wordArray[i] > wordArray[j])
{
x++;
}
}
rank = rank + x * (Factorial(n - i - 1));
}
return rank;
}
//Ranks the word input if there are repeated charaters
//in the word
static Int64 rankPerm(String word)
{
Int64 rank = 1;
Int64 suffixPermCount = 1;
SortedDictionary<char, int> counter = Counter(word);
for (int i = word.Length - 1; i > -1; i--)
{
char x = Convert.ToChar(word.Substring(i,1));
int xCount;
if(counter[x] != 0)
{
xCount = counter[x] + 1;
}
else
{
xCount = 1;
}
counter[x] = xCount;
foreach (KeyValuePair<char,int> e in counter)
{
if (e.Key < x)
{
rank += suffixPermCount * e.Value / xCount;
}
}
suffixPermCount *= word.Length - i;
suffixPermCount /= xCount;
}
return rank;
}
static void Main(string[] args)
{
Console.WriteLine("Type Exit to end the program.");
string prompt = "Please enter a word using only letters:";
const int MAX_VALUE = 25;
Int64 rank = new Int64();
string theWord;
do
{
theWord = readWord(prompt, MAX_VALUE);
char[] wordLetters = theWord.ToCharArray();
Array.Sort(wordLetters);
bool duplicate = false;
for(int i = 0; i< theWord.Length - 1; i++)
{
if(wordLetters[i] < wordLetters[i+1])
{
duplicate = true;
}
}
if(duplicate)
{
SortedDictionary<char, int> counter = Counter(theWord);
rank = rankPerm(theWord);
Console.WriteLine("\n" + theWord + " = " + rank);
}
else
{
char[] letters = theWord.ToCharArray();
rank = rankWord(letters);
Console.WriteLine("\n" + theWord + " = " + rank);
}
} while (theWord != "EXIT");
Console.WriteLine("\nPress enter to escape..");
Console.Read();
}
}
}
If there are k distinct characters, the i^th character repeated n_i times, then the total number of permutations is given by
(n_1 + n_2 + ..+ n_k)!
------------------------------------------------
n_1! n_2! ... n_k!
which is the multinomial coefficient.
Now we can use this to compute the rank of a given permutation as follows:
Consider the first character(leftmost). say it was the r^th one in the sorted order of characters.
Now if you replace the first character by any of the 1,2,3,..,(r-1)^th character and consider all possible permutations, each of these permutations will precede the given permutation. The total number can be computed using the above formula.
Once you compute the number for the first character, fix the first character, and repeat the same with the second character and so on.
Here's the C++ implementation to your question
#include<iostream>
using namespace std;
int fact(int f) {
if (f == 0) return 1;
if (f <= 2) return f;
return (f * fact(f - 1));
}
int solve(string s,int n) {
int ans = 1;
int arr[26] = {0};
int len = n - 1;
for (int i = 0; i < n; i++) {
s[i] = toupper(s[i]);
arr[s[i] - 'A']++;
}
for(int i = 0; i < n; i++) {
int temp = 0;
int x = 1;
char c = s[i];
for(int j = 0; j < c - 'A'; j++) temp += arr[j];
for (int j = 0; j < 26; j++) x = (x * fact(arr[j]));
arr[c - 'A']--;
ans = ans + (temp * ((fact(len)) / x));
len--;
}
return ans;
}
int main() {
int i,n;
string s;
cin>>s;
n=s.size();
cout << solve(s,n);
return 0;
}
Java version of unrank for a String:
public static String unrankperm(String letters, int rank) {
Map<Character, Integer> charCounts = new java.util.HashMap<>();
int permcount = 1;
for(int i = 0; i < letters.length(); i++) {
char x = letters.charAt(i);
int xCount = charCounts.containsKey(x) ? charCounts.get(x) + 1 : 1;
charCounts.put(x, xCount);
permcount = (permcount * (i + 1)) / xCount;
}
// charCounts is the histogram of letters
// permcount is the number of distinct perms of letters
StringBuilder perm = new StringBuilder();
for(int i = 0; i < letters.length(); i++) {
List<Character> sorted = new ArrayList<>(charCounts.keySet());
Collections.sort(sorted);
for(Character x : sorted) {
// suffixcount is the number of distinct perms that begin with x
Integer frequency = charCounts.get(x);
int suffixcount = permcount * frequency / (letters.length() - i);
if (rank <= suffixcount) {
perm.append(x);
permcount = suffixcount;
if(frequency == 1) {
charCounts.remove(x);
} else {
charCounts.put(x, frequency - 1);
}
break;
}
rank -= suffixcount;
}
}
return perm.toString();
}
See also n-th-permutation-algorithm-for-use-in-brute-force-bin-packaging-parallelization.
I looked at http://rosettacode.org/wiki/Knapsack_problem/0-1 to do the basic knapsack dynamic programming problem, and I got a working solution (knapsack1()), but when I tried a different solution (knapsack2()) I feel like I am off-by-one somewhere because I am not getting the correct value.
The code in question is the method knapsack2, at the bottom. I'm sure the issue is small, but I feel absurd because I can't find the problem. The correct answer should be 1030 and not 880.
(I know that an array of size [n][S] should be enough, as shown in knapsack1(), which works, but knapsack2() doesn't oddly.)
Thanks in advance for anyone who looks through this.
/**
* http://rosettacode.org/wiki/Knapsack_problem/0-1
*/
public class Knapsack {
public static void main(String[] args) {
int n = 22;
int S = 400;
int s[] = new int[22];
int v[] = new int[22];
int i = 0;
s[i] = 9;
v[i] = 150;
i++;
s[i] = 13;
v[i] = 35;
i++;
s[i] = 153;
v[i] = 200;
i++;
s[i] = 50;
v[i] = 160;
i++;
s[i] = 15;
v[i] = 60;
i++;
s[i] = 68;
v[i] = 45;
i++;
s[i] = 27;
v[i] = 60;
i++;
s[i] = 39;
v[i] = 40;
i++;
s[i] = 23;
v[i] = 30;
i++;
s[i] = 52;
v[i] = 10;
i++;
s[i] = 11;
v[i] = 70;
i++;
s[i] = 32;
v[i] = 30;
i++;
s[i] = 24;
v[i] = 15;
i++;
s[i] = 48;
v[i] = 10;
i++;
s[i] = 73;
v[i] = 40;
i++;
s[i] = 42;
v[i] = 70;
i++;
s[i] = 43;
v[i] = 75;
i++;
s[i] = 22;
v[i] = 80;
i++;
s[i] = 7;
v[i] = 20;
i++;
s[i] = 18;
v[i] = 12;
i++;
s[i] = 4;
v[i] = 50;
i++;
s[i] = 30;
v[i] = 10;
System.out.println("--Given items with these values--");
System.out.println("[item, weight (dag), value]");
System.out.println("map 9 150");
System.out.println("compass 13 35");
System.out.println("water 153 200");
System.out.println("sandwich 50 160");
System.out.println("glucose 15 60");
System.out.println("tin 68 45");
System.out.println("banana 27 60");
System.out.println("apple 39 40");
System.out.println("cheese 23 30");
System.out.println("beer 52 10");
System.out.println("suntan cream 11 70");
System.out.println("camera 32 30");
System.out.println("T-shirt 24 15");
System.out.println("trousers 48 10");
System.out.println("umbrella 73 40");
System.out.println("waterproof trousers 42 70");
System.out.println("waterproof overclothes 43 75");
System.out.println("note-case 22 80");
System.out.println("sunglasses 7 20");
System.out.println("towel 18 12");
System.out.println("socks 4 50");
System.out.println("book 30 10");
System.out.println("--The max value you can achieve in your knapsack--");
System.out.println(knapsack2(n, S, s, v));
// proper value should be 1030, from website
}
/**
* #param n items
* #param S bag size/capacity
* #param s array where s[i] is size/weight of item at index i
* #param v array where v[i] is value of item at index i
* #return best/max value possible
*/
#SuppressWarnings("unused")
private static int knapsack1(int n, int S, int s[], int v[]) {
int dp[][] = new int[n][S];
for (int i=n-1; i >= 0; i--) { // # items being left out + 1
for (int j=0; j < S; j++) { // size/weight + 1
if (i == n-1) {
dp[i][j] = 0;
} else {
int choices[] = {0,0};
choices[0] = dp[i+1][j];
if (j >= s[i])
choices[1] = v[i] + dp[i+1][j-s[i]];
dp[i][j] = max(choices);
}
}
}
return dp[0][S-1];
}
private static int max(int choices[]) {
if (choices[0] > choices[1])
return choices[0];
else
return choices[1];
}
/**
* #param n items
* #param S bag size/capacity
* #param s array where s[i] is size/weight of item at index i
* #param v array where v[i] is value of item at index i
* #return best/max value possible
*/
// #SuppressWarnings("unused")
private static int knapsack2(int n, int S, int s[], int v[]) {
int dp[][] = new int[n][S]; // dp[i][j] holds max value using i items and not exceeding weight j+1
for (int i=0; i < n; i++) { // # items. don't need to reach n because we assume we can fit at most n-1 items.
for (int j=0; j < S; j++) { // size/weight+1
if (i == 0) {
dp[i][j] = 0;
} else {
int choices[] = {0,0};
choices[0] = dp[i-1][j];
if (j >= s[i])
choices[1] = v[i] + dp[i-1][j-s[i]];
dp[i][j] = max(choices);
}
}
}
return dp[n-1][S-1]; // TODO: WHY IS THIS 880 AND NOT 1030?
}
}
Both your knapsack functions have issues. Both won't pass the simple test - S=1 n=1 s[0]=1 v[0]=10. The answer should be 10 but your code will return 0.
You can see that the issue are in the lines if (i==(n-1)) (knapsack1) or if (i==0) (knapsack2).
Also, you will see issues because of condition if (j >= s[i]. Your j starts at 0 so, if S was 1, you'll never be able to the item with s[i] == 1 because j will never reach 1.
Hope this helps. Good luck.