LightOJ - 1051 Good or Bad（扫描）

题目大意：给你一个字符串，如果出现三个相邻的辅音字母，或者三个相邻的原音字母，这个字符串就是BAD的，反之就是good。字符串中有？，允许将？变成26个字母中的任意一个。如果字符串能变化成BAD和GOOD，那么就输出MIXED

解题思路：先不考虑？，看是否是BAD，如果是BAD就不用再考虑了

再考虑一下所有？变成原因和辅音的两种情况，看是否会变成BAD

最后考虑一下，是否永远都是BAD，如果不是永远都是BAD，就表示字符串可以变GOOD

判断是否永远是BAD，只要判断一下是否?无论怎么变都是bad就可以

#include <cstdio>
#include <cstring>
const int N = 60;

char str
;
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 (rightCon) 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;
}