解决密码算法难题

2024-04-22 20:10:10 20

在隐式算术问题中，一些字母用于为其分配数字。就像十个不同的字母一样，它们保持从0到9的数字值以正确执行算术运算。给了两个单词，另一个单词给出了这两个单词的加法答案。

例如，我们可以说两个单词“BASE”和“BALL”，结果为“GAMES”。现在，如果我们尝试通过其符号数字添加BASE和BALL，我们将得到答案GAMES。

注意＆minuns; 最多不能超过十个字母，否则无法解决。

输入输出

Input:
This algorithm will take three words.
             B A S E
             B A L L
           ----------
           G A M E S

Output:
It will show which letter holds which number from 0 – 9.
For this case it is like this.              B A S E                         2 4 6 1
              B A L L                         2 4 5 5
             ---------                       ---------
            G A M E S                       0 4 9 1 6

算法

对于此问题，我们将定义一个节点，其中包含一个字母及其对应的值。

isValid(nodeList,count,word1,word2,word3)

输入-节点列表，节点列表中的元素数和三个单词。

输出-如果word1和word2的值之和与word3的值相同，则为True。

Begin
   m := 1
   for each letter i from right to left of word1, do
      ch := word1[i]
      for all elements j in the nodeList, do
         if nodeList[j].letter = ch, then
            break
      done
      val1 := val1 + (m * nodeList[j].value)
      m := m * 10
   done

   m := 1
   for each letter i from right to left of word2, do
      ch := word2[i]
      for all elements j in the nodeList, do
         if nodeList[j].letter = ch, then
            break
      done

      val2 := val2 + (m * nodeList[j].value)
      m := m * 10
   done

   m := 1
   for each letter i from right to left of word3, do
      ch := word3[i]
      for all elements j in the nodeList, do
         if nodeList[j].letter = ch, then
            break
      done

      val3 := val3 + (m * nodeList[j].value)
      m := m * 10
   done

   if val3 = (val1 + val2), then
      return true
   return false
End

排列(nodeList,count,n,word1,word2,word3)

输入-节点列表，列表中的项目数，分配的字母和三个单词的数量。

输出-为所有字母正确分配值以求和时为True。

Begin
   if n letters are assigned, then
      for all digits i from 0 to 9, do
         if digit i is not used, then
            nodeList[n].value := i
            if isValid(nodeList, count, word1, word2, word3) = true
               for all items j in the nodeList, do
                  show the letter and corresponding values.
               done
               return true
      done
      return fasle

   for all digits i from 0 to 9, do
      if digit i is not used, then
         nodeList[n].value := i
         mark as i is used
         if permutation(nodeList, count, n+1, word1, word2, word3),
            return true
         otherwise mark i as not used
   done
   return false
End

示例

#include <iostream>
#include<vector>
using namespace std;

vector<int> use(10);        //set 1, when one character is assigned previously
struct node {
   char letter;
   int value;
};

int isValid(node* nodeList, const int count, string s1, string s2, string s3) {
   int val1 = 0, val2 = 0, val3 = 0, m = 1, j, i;

   for (i = s1.length() - 1; i >= 0; i--) {    //find number for first string
      char ch = s1[i];
      for (j = 0; j < count; j++)
         if (nodeList[j].letter == ch)       //when ch is present, break the loop
            break;
      val1 += m * nodeList[j].value;
      m *= 10;
   }

   m = 1;
   for (i = s2.length() - 1; i >= 0; i--) {    //find number for second string
      char ch = s2[i];
      for (j = 0; j < count; j++)
         if (nodeList[j].letter == ch)
            break;
      val2 += m * nodeList[j].value;
      m *= 10;
   }

   m = 1;
   for (i = s3.length() - 1; i >= 0; i--) {    //find number for third string
      char ch = s3[i];
      for (j = 0; j < count; j++)
         if (nodeList[j].letter == ch)
            break;
      val3 += m * nodeList[j].value;
      m *= 10;
   }

   if (val3 == (val1 + val2))    //check whether the sum is same as 3rd string or not
      return 1;
   return 0;
}

bool permutation(int count, node* nodeList, int n, string s1, string s2, string s3) {
   if (n == count - 1) {     //when values are assigned for all characters
      for (int i = 0; i < 10; i++) {
         if (use[i] == 0) {     // for those numbers, which are not used
            nodeList[n].value = i;    //assign value i
            if (isValid(nodeList, count, s1, s2, s3) == 1) { //check validation
               cout << "Solution found: ";
               for (int j = 0; j < count; j++)    //print code, which are assigned
                  cout << " " << nodeList[j].letter << " = " << nodeList[j].value;
               return true;
            }
         }
      }
      return false;
   }

   for (int i = 0; i < 10; i++) {
      if (use[i] == 0) {           // for those numbers, which are not used
         nodeList[n].value = i;    //assign value i and mark as not available for future use
         use[i] = 1;
         if (permutation(count, nodeList, n + 1, s1, s2, s3))    //go for next characters
            return true;
         use[i] = 0; //when backtracks, make available again
      }
   }
   return false;
}

bool solvePuzzle(string s1, string s2,string s3) {
   int uniqueChar = 0; //Number of unique characters
   int len1 = s1.length();
   int len2 = s2.length();
   int len3 = s3.length();

   vector<int> freq(26); //There are 26 different characters

   for (int i = 0; i < len1; i++)
      ++freq[s1[i] - 'A'];
   for (int i = 0; i < len2; i++)
      ++freq[s2[i] - 'A'];
   for (int i = 0; i < len3; i++)
      ++freq[s3[i] - 'A'];

   for (int i = 0; i < 26; i++)
      if (freq[i] > 0)     //whose frequency is > 0, they are present
         uniqueChar++;

   if (uniqueChar > 10) { //as there are 10 digits in decimal system
      cout << "Invalid strings";
      return 0;
   }

   node nodeList[uniqueChar];
   for (int i = 0, j = 0; i < 26; i++) {     //assign all characters found in three strings
      if (freq[i] > 0) {
         nodeList[j].letter = char(i + 'A');
         j++;
      }
   }
   return permutation(uniqueChar, nodeList, 0, s1, s2, s3);
}

int main() {
   string s1 = "BASE";
   string s2 = "BALL";
   string s3 = "GAMES";

   if (solvePuzzle(s1, s2, s3) == false)
      cout << "No solution";
}

输出结果

Solution found:  A = 4 B = 2 E = 1 G = 0 L = 5 M = 9 S = 6

解决密码算法难题

输入输出

算法

示例

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