1940-回文对称

 1

 #include
 2 char s1[10005], s2[10005];
 3 void judge() {
 4     int len2 = 0;
 5     for (int j = 0; s1[j] != '\0'; j++) {
 6         len2++;
 7         if (s1[j] == '(') s2[j] = s1[j] + 1;
 8         else s2[j] = s1[j] - 1;
 9         s2[j + 1] = '\0';
10     }
11     int k;
12     for ( k = len2 - 1; k >= 0; k--) {
13         if (s2[k] != s1[len2 - k - 1]) {//REVERSE
14             break;
15         }
16     }if (k == -1) {
17         printf("SYMMETRY!!!\n");
18         return;
19     }int l;
20     for ( l = len2 - 1; l >= 0; l--) {
21         if (s1[l] != s1[len2 - l - 1]) {
22             break;
23         }
24     }if (l == -1) {
25         printf("PALINDROME!!!\n");
26         return;
27     }
28     printf("UGLY!!!\n");
29 }
30 
31 int main() {
32     int t;
33     scanf("%d", &t);
34     for (int i = 0; i < t; i++) {
35         scanf("%s", s1);
36         judge();
37     }
38 }