Uvs: 185-Roman Numerals (dfs)

Question: Ultraviolet A: 185-Roman Numerals

Question: Give a string equality and ask if this string is a Rome equality? Are there any conforming Arabic equations? The former outputs correct or incorrect, and the latter scores:

A combination of letters that can form an Arabic equation must be greater than or equal to 2,
There is only one combination of valid letters that can form an Arabic equation.
Impossible does not have a combination of letters that match the Arabic equation.
Solution:
1. Rule of making up the Rome equation: if the value of the letter (I-1) on the left of each current letter (I) is smaller than the value corresponding to I, the value of the I-1 is negative, otherwise it is positive. Finally, determine whether or not it is just about or not.
2. If an Arabic equation can be formed, each letter can be replaced by a number between 0 and 9 to see if there is an equation between the left and right sides.
Note:
0 is displayed separately, and 0 is not used in the case of leading 0, and must be excluded. Then the same letter indicates the same value. Here we will enumerate the situations of each letter, and then remove the case where 0 is wrong, and dfs () will find the correct combination.

Code:
#include 
   
    #include 
    
     const int N = 10;char str[N * N], s1[3][N];int s[N * N];char vis[N * N], head[N * N];const char *letter = "IVXLCDM";intcount, size;void init () {s['I'] = 1;s['X'] = 10;s['C'] = 100;s['M'] = 1000;s['V'] = 5;s['L'] = 50;s['D'] = 500;size = 0;}int trans (char * a) {int num = 0, i;for (i = 1; i <= strlen (a); i++) {if (s[a[i - 1]] >= s[a[i]])num += s[a[i- 1]];elsenum -= s[a[i - 1]];}return num;}bool is_roman () {int n1 = trans(s1[0]);int n2 = trans(s1[1]);int n3 = trans(s1[2]);if (n1 + n2 == n3)return true;return false;}int tran_num (char * a) {int num = 0;for (int i = 0; i < strlen (a); i++)num = num * 10 + s[a[i]];return num;}void is_numerals (int n, int v) {if (n == size) {int n1 = tran_num (s1[0]);int n2 = tran_num (s1[1]);int n3 = tran_num (s1[2]);if (n1 + n2 == n3) count++;return ;}if (v >= 7)return;while(!vis[letter[v]]) v++;if (v < 7) {for (int j = 0; j < 10; j++) {if (j == 0 && head[letter[v]])continue;s[letter[v]] = j;is_numerals (n + 1, v + 1);if (count >= 2)return;}} }int main () {int k, j;while (scanf ("%s", str) , str[0] != '#') {init();k = j = 0;memset (vis, 0, sizeof (vis));memset (head, 0, sizeof (head));for (int i = 0; i < strlen (str); i++) {if (str[i] != '+' && str[i] != '=' ) {s1[k][j++] = str[i];if (!vis[str[i]])size++;vis[str[i]] = 1;if (j == 1)head[str[i]] = 1;} else {s1[k][j] = '\0';k++;j = 0;}}s1[2][j] = '\0';printf ("%s ", is_roman() ?"Correct":"Incorrect");count = 0;is_numerals(0, 0);if (!count)printf ("impossible\n");else {if (count == 1)printf ("valid\n");elseprintf ("ambiguous\n");}}return 0;}