报纸和杂志通常具有以下形式的隐式算术难题：
例子：
Input : s1 = SEND, s2 = "MORE", s3 = "MONEY"
Output : One of the possible solution is:
D=1 E=5 M=0 N=3 O=8 R=2 S=7 Y=6
Explanation:
The above values satisfy below equation :
SEND
+ MORE
--------
MONEY
--------
强烈建议参考Backtracking |设置8(解决密码学难题)来解决这个问题。
这个想法是给每个字母分配一个从0到9的数字，以便算术正确地进行计算。排列是一种递归函数，它针对整数的每个可能排列调用检查函数。
Check函数检查与前两个字符串相对应的前两个数字的和是否等于与第三个字符串相对应的第三个数字的和。如果找到解决方案，则打印解决方案。
// CPP program for solving cryptographic puzzles
#include
using namespace std;
// vector stores 1 corresponding to index
// number which is already assigned
// to any char, otherwise stores 0
vector use(10);
// structure to store char and its corresponding integer
struct node
{
char c;
int v;
};
// function check for correct solution
int check(node* nodeArr, const int count, string s1,
string s2, string s3)
{
int val1 = 0, val2 = 0, val3 = 0, m = 1, j, i;
// calculate number corresponding to first string
for (i = s1.length() - 1; i >= 0; i--)
{
char ch = s1[i];
for (j = 0; j < count; j++)
if (nodeArr[j].c == ch)
break;
val1 += m * nodeArr[j].v;
m *= 10;
}
m = 1;
// calculate number corresponding to second string
for (i = s2.length() - 1; i >= 0; i--)
{
char ch = s2[i];
for (j = 0; j < count; j++)
if (nodeArr[j].c == ch)
break;
val2 += m * nodeArr[j].v;
m *= 10;
}
m = 1;
// calculate number corresponding to third string
for (i = s3.length() - 1; i >= 0; i--)
{
char ch = s3[i];
for (j = 0; j < count; j++)
if (nodeArr[j].c == ch)
break;
val3 += m * nodeArr[j].v;
m *= 10;
}
// sum of first two number equal to third return true
if (val3 == (val1 + val2))
return 1;
// else return false
return 0;
}
// Recursive function to check solution for all permutations
bool permutation(const int count, node* nodeArr, int n,
string s1, string s2, string s3)
{
// Base case
if (n == count - 1)
{
// check for all numbers not used yet
for (int i = 0; i < 10; i++)
{
// if not used
if (use[i] == 0)
{
// assign char at index n integer i
nodeArr[n].v = i;
// if solution found
if (check(nodeArr, count, s1, s2, s3) == 1)
{
cout << "\nSolution found: ";
for (int j = 0; j < count; j++)
cout << " " << nodeArr[j].c << " = "
<< nodeArr[j].v;
return true;
}
}
}
return false;
}
for (int i = 0; i < 10; i++)
{
// if ith integer not used yet
if (use[i] == 0)
{
// assign char at index n integer i
nodeArr[n].v = i;
// mark it as not available for other char
use[i] = 1;
// call recursive function
if (permutation(count, nodeArr, n + 1, s1, s2, s3))
return true;
// backtrack for all other possible solutions
use[i] = 0;
}
}
return false;
}
bool solveCryptographic(string s1, string s2,
string s3)
{
// count to store number of unique char
int count = 0;
// Length of all three strings
int l1 = s1.length();
int l2 = s2.length();
int l3 = s3.length();
// vector to store frequency of each char
vector freq(26);
for (int i = 0; i < l1; i++)
++freq[s1[i] - 'A'];
for (int i = 0; i < l2; i++)
++freq[s2[i] - 'A'];
for (int i = 0; i < l3; i++)
++freq[s3[i] - 'A'];
// count number of unique char
for (int i = 0; i < 26; i++)
if (freq[i] > 0)
count++;
// solution not possible for count greater than 10
if (count > 10)
{
cout << "Invalid strings";
return 0;
}
// array of nodes
node nodeArr[count];
// store all unique char in nodeArr
for (int i = 0, j = 0; i < 26; i++)
{
if (freq[i] > 0)
{
nodeArr[j].c = char(i + 'A');
j++;
}
}
return permutation(count, nodeArr, 0, s1, s2, s3);
}
// Driver function
int main()
{
string s1 = "SEND";
string s2 = "MORE";
string s3 = "MONEY";
if (solveCryptographic(s1, s2, s3) == false)
cout << "No solution";
return 0;
}
输出：
Solution found: D=1 E=5 M=0 N=3 O=8 R=2 S=7 Y=6
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