MU Puzzle HDU - 4662

Suppose there are the symbols M, I, and U which can be combined to produce strings of symbols called "words". We start with one word MI, and transform it to get a new word. In each step, we can use one of the following transformation rules: 
1. Double any string after the M (that is, change Mx, to Mxx). For example: MIU to MIUIU. 
2. Replace any III with a U. For example: MUIIIU to MUUU. 
3. Remove any UU. For example: MUUU to MU. 
Using these three rules is it possible to change MI into a given string in a finite number of steps?

InputFirst line, number of strings, n. 
Following n lines, each line contains a nonempty string which consists only of letters ‘M‘, ‘I‘ and ‘U‘.

Total length of all strings <= 10 ⁶.Outputn lines, each line is ‘Yes‘ or ‘No‘.Sample Input

2
MI
MU

Sample Output

Yes
No

所有U都是由I换来的，而U不能再换成I，所以可以讲所有U换成I的数目进行统计，如果符合要求即可1.相当于I的数目*23.相当于I的数目-6

#include <cstdio>
#include <cstring>

using namespace std;

const int maxn = 1000000 + 10;
char s[maxn];
int p[30];

void init(){
    p[0] = 1;
    for(int i = 1; i < 30; i++) p[i] = (p[i-1] << 1);
}

bool solve(){
    bool ok = 0;
    int len = strlen(s), i, Mcnt = 0, Icnt = 0;
    for(i = 0; i < len; i++){
        if(s[i] == ‘M‘) Mcnt++;
        else if(s[i] == ‘I‘) Icnt++;
        else if(s[i] == ‘U‘) Icnt += 3;
    }
    if(Mcnt == 1 && s[0] == ‘M‘){
        for(i = 29; i >= 0; i--) if(p[i] >= Icnt && (p[i] - Icnt) % 6 == 0) {
            ok = 1;
            break;
        }
    }
    return ok;
}

int main()
{
    int n;
    init();
    scanf("%d", &n);
    getchar();
    while(n--){
        scanf("%s", s);
        if(solve()) printf("Yes\n");
        else printf("No\n");
    }
    return 0;
}