hdu 4662 MU Puzzle

Problem Description

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 <= 10⁶.

 

Output

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

 

Sample Input

2MIMU

Sample Output

YesNo
1、Mx -> Mxx
2、III -> U
3、UU -> 空
问MI 能否进过上面操作转化成  读入的字符串。
重点是怎么推式子
1、M前面的不能有字母。
2、只能有一个M
3、操作一只能得到 2^n个I (n > 0)
4、可以消去6个I.
然后得到一个公式  2^n - 6 * m = sum; (sum 表示读入的字符串中I的个数);
#include<iostream>#include<cstring>using namespace std;char s[1000001];int main(){    int n,j,u,i;    cin>>n;    while(n--)    {        cin>>s;        u=i=0;        if(s[0]!='M'||s[1]==0)        {            cout<<"No"<<endl;        }        else if(strcmp(s,"MI")==0)        {            cout<<"Yes"<<endl;        }        else if(strcmp(s,"MU")==0)        {            cout<<"No"<<endl;        }        else        {            for(j=1;s[j];j++)            {                if(s[j]=='U')                u++;                else if(s[j]=='I')                i++;                if(s[j]=='M')                {                    cout<<"No"<<endl;                    break;                }            }            if(s[j])            {                continue;            }            if((u*3+i)%2==0&&i>0&&(u*3+i)%3!=0)            cout<<"Yes"<<endl;            else            cout<<"No"<<endl;        }    }    return 0;}