1940-回文对称

 #include<bits/stdc++.h>
char s1[10005], s2[10005];
void judge() {
    int len2 = 0;
    for (int j = 0; s1[j] != '\0'; j++) {
        len2++;
        if (s1[j] == '(') s2[j] = s1[j] + 1;
        else s2[j] = s1[j] - 1;
        s2[j + 1] = '\0';
    }
    int k;
    for ( k = len2 - 1; k >= 0; k--) {
        if (s2[k] != s1[len2 - k - 1]) {//REVERSE
            break;
        }
    }if (k == -1) {
        printf("SYMMETRY!!!\n");
        return;
    }int l;
    for ( l = len2 - 1; l >= 0; l--) {
        if (s1[l] != s1[len2 - l - 1]) {
            break;
        }
    }if (l == -1) {
        printf("PALINDROME!!!\n");
        return;
    }
    printf("UGLY!!!\n");
}

int main() {
    int t;
    scanf("%d", &t);
    for (int i = 0; i < t; i++) {
        scanf("%s", s1);
        judge();
    }
}