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。
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
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
#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