C++程序解决算术谜题
在算术谜题中,一些字母被用来赋值。例如,十个不同的字母持有从0到9的数字值,以正确执行算术运算。给出两个词,另一个词作为这两个词加法的答案。例如,两个词“BASE”和“BALL”,结果是“GAMES”。如果我们尝试用它们的符号数字来添加BASE和BALL,我们将得到答案GAMES。
注意 - 最多必须有十个字母,否则无法求解。
输入
此算法将接收三个单词。
输出
它将显示哪个字母持有0-9中的哪个数字。
对于这种情况,是这样的。
字母 | A | B | E | G | L | M | S |
值 | 4 | 2 | 1 | 0 | 5 | 9 | 6 |
算法
对于这个问题,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,节点列表中的元素个数和三个单词。
输出 - 如果word1和word2的值之和与word3的值相同,则返回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)
输入 - 节点列表,列表中的项目数,已赋值字母的数量和三个单词。
输出 - 当所有字母都被正确赋值以解决求和问题时返回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
#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
Solution found: A = 4 B = 2 E = 1 G = 0 L = 5 M = 9 S = 6
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