HDU - 4662 MU Puzzle

版权声明：转载时标明一下出处啦QwQ https://blog.csdn.net/Zero_979/article/details/81539933

题目地址：http://acm.hdu.edu.cn/showproblem.php?pid=4662

题目：

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? 

Input

First 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.

Output

n lines, each line is 'Yes' or 'No'.

Sample Input

2
MI
MU

Sample Output

Yes
No

思路：

因为一个U等于三个I，所以可以把U都统计到I上，然后由于M后的字母可以扩大两倍，所以MI可以有的I是1,2,4,8……都是2的幂。而两个U可以消除，也就是说6个I可以消除，那么只要用比I大的2的幂-I去模6等于0就可以变成MI。

代码：

#include <cstdio>
#include <cstring>
#include<iostream>
using namespace std;
const int maxn = 1000010;
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, M = 0, I = 0;
    for(i = 0; i < len; i++)
    {
        if(s[i] == 'M') M++;
        else if(s[i] == 'I') I++;
        else if(s[i] == 'U') I += 3;
    }
    if(M == 1 && s[0] == 'M')
    {
        for(i = 29; i >= 0; i--)
            if(p[i] >= I && (p[i] - I) % 6 == 0) { ok = 1; break; }
    }
    return ok;
}
int main()
{   int n;
    init();
    cin>>n;
    getchar();
    while(n--)
    {
        scanf("%s", s);
        if(solve()) cout<<"Yes"<<endl;
        else cout<<"No"<<endl;
    }
    return 0;
}