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
题意：判断给定的字符串能否由MI通过3种规则转换而来。
题解：通过对3种规则的联系，发现全部U可以转换为I，即每个U换3个I。
思路：把全部的U换成I，看I的个数是否为2的幂，由规则2可以知 I 的个数可以减6再判断是不是2的幂。
坑点：1.要判断M是不只有是一个
2.要判断M是不是在第一位
3.要注意MI这种要单独判断，且 I 的数量减6之后不能把I=1的情况算进去，要单独考虑
技巧：判断2的幂的简单写法：
bool isPowerOf2(int n){
if(n <= 0) return false;
else return (n & (n-1)) == 0;
}
贴代码：
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <cstring>
using namespace std;
bool isPowerOf2(int n){
if(n <= 0){
return false;
}
return(n & (n-1)) == 0;
}
int a[1000005];
char s[1000005];
int main()
{
int n;
cin>>n;
while(n--){
cin>>s;
int len = strlen(s);
int m = 0,u = 0,i = 0;
if(s[0] != 'M') {cout<<"No"<<endl;continue;}
for(int k = 0;k < len;k++){
if(s[k] == 'M') m++;
else if(s[k] == 'U') u++;
else if(s[k] == 'I') i++;
}
if(m > 1) {cout<<"No"<<endl;continue;}
if(u==0 && i==1) {cout<<"Yes"<<endl;continue;}
i += u*3;
bool flag = true;
while(i > 1){
if(isPowerOf2(i)) {cout<<"Yes"<<endl;flag = false;break;}
else i-=6;
}
if(flag) cout<<"No"<<endl;
}
return 0;
}