C++程式解決算術謎題


在算術謎題中,一些字母被用來賦值。例如,十個不同的字母持有從0到9的數字值,以正確執行算術運算。給出兩個詞,另一個詞作為這兩個詞加法的答案。例如,兩個詞“BASE”和“BALL”,結果是“GAMES”。如果我們嘗試用它們的符號數字來新增BASE和BALL,我們將得到答案GAMES。

注意 - 最多必須有十個字母,否則無法求解。

輸入

此演算法將接收三個單詞。

輸出

它將顯示哪個字母持有0-9中的哪個數字。

對於這種情況,是這樣的。

字母ABEGLMS
4210596


 

演算法

對於這個問題,我們將定義一個節點,其中包含一個字母及其對應的值。

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

輸入 - 節點列表,節點列表中的元素個數和三個單詞。
輸出 - 如果word1和word2的值之和與word3的值相同,則返回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

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

輸入 - 節點列表,列表中的專案數,已賦值字母的數量和三個單詞。
輸出 - 當所有字母都被正確賦值以解決求和問題時返回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

線上演示

#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";
}

輸出

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

更新於: 2019年7月30日

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