LightOJ-1051 good or bad (scan)

Title: Give you a string, if there are three adjacent consonants, or three adjacent original letters, the string is bad, and vice versa is good. There are in the string. , allowing the. into any of the 26 letters. If the string can change to bad and good, then the output mixed

Problem-solving ideas: Do not consider first. To see if it's bad, and if it's bad, you don't have to think about it.
Think about everything again. Become a cause and a consonant in two cases, see if it turns bad
Finally, consider whether it is always bad, if not always, it means that the string can be changed good
Judging whether it is bad forever, just judge it? No matter how bad it's going to be.

#include <cstdio> #include <cstring> const int N = 60;
Char Str[n];

int len, cas = 1;
    void Init () {scanf ("%s", str);
len = strlen (str);
    } void Solve () {int vow = 0, con = 0;
    BOOL Bad = 0, good = 0, Allbad = 0; for (int i = 0; i < len; i++) {if (str[i] = = ' A ' | | str[i] = = ' E ' | | str[i] = = ' I ' | | str[i] = = ' O ' | | str[i]
            = = ' U ') {vow++;
        con = 0; } else if (str[i] = = '? ')
            {vow = 0;
        con = 0;
            } else {con++;
        Vow = 0;
            } if (vow = = 3 | | con = = 5) {Allbad = bad = 1;
        Break
        }} if (Allbad) {printf ("Case%d:bad\n", cas++);
    return;
        } if (!bad) {vow = 0, con = 0; for (int i = 0; i < len; i++) {if (str[i] = = ' A ' | | str[i] = = ' E ' | | str[i] = = ' I ' | | str[i] = = ' O ' | | | str [i] = = ' U ' | | Str[i] = = '? ')
       {vow++;         con = 0;
                } else {vow = 0;
            con++;
                } if (vow = = 3) {bad = true;
            Break
        }}} if (!bad) {vow = 0, con = 0; for (int i = 0; i < len; i++) {if (str[i] = = ' A ' | | str[i] = = ' E ' | | str[i] = = ' I ' | | str[i] = = ' O ' | | | str
                [i] = = ' U ') {vow++;
            con = 0;
                } else {con++;
            Vow = 0;
                } if (con = = 5) {bad = true;
            Break }}} for (int i = 0; i < len; i++) {if (str[i] = = '? ')
            {int leftvow = 0, Leftcon = 0, rightvow = 0, Rightcon = 0;
                for (int j = i-1; J >= 0, j--) {if (str[j] = = '? ') break;
   if (str[j] = = ' A ' | | str[j] = = ' E ' | | str[j] = = ' I ' | | str[j] = = ' O ' | | str[j] = = ' U ') {                 if (Leftcon) break;
                leftvow++;
                    } else {if (leftvow) break;
                leftcon++;
                }} for (int j = i + 1, j < Len; J + +) {if (str[j] = = '? ') break; if (str[j] = = ' A ' | | str[j] = = ' E ' | | str[j] = = ' I ' | | str[j] = = ' O ' | | str[j] = = ' U ') {if (rig
                    Htcon) break;
                rightvow++;
                    } else {if (rightvow) break;
                rightcon++; }} if ((Leftvow = = 2 && Rightcon = = 4) | | (Leftcon = = 4 && rightvow = = 2))
            Allbad = 1;
            if (leftvow = = 2 | | rightvow = = 2) str[i] = ' Q ';
            if (Leftcon = = 4 | | rightcon = = 4) str[i] = ' A ';
        if (Allbad) break;
    }} if (Allbad) printf ("Case%d:bad\n", cas++); else if (bad) printf (' Case%d:miXed\n ", cas++);
else if (!bad) printf ("Case%d:good\n", cas++);
    } int main () {int test;
    scanf ("%d", &test);
        while (test--) {init ();
    Solve ();
} return 0;
 }