hdu 4622 **

题意: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?

首先全部转化为I

可以发现，能被6整除或者i的个数是奇数且不为1的不可以，其他都可以

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<queue>
#include<map>
using namespace std;
#define MOD 1000000007
const int INF=0x3f3f3f3f;
const double eps=1e-5;
typedef long long ll;
#define cl(a) memset(a,0,sizeof(a))
#define ts printf("*****\n");
const int MAXN=1005055;
int n,m,tt;
char s[MAXN];
int main()
{
    int i,j,k;
    #ifndef ONLINE_JUDGE
    freopen("1.in","r",stdin);
    #endif
    scanf("%d",&tt);
    while(tt--)
    {
        scanf("%s",s);
        if(s[0]!='M')
        {
            printf("No\n");
            continue;
        }
        int cnt=0;
        bool flag=1;
        int len=strlen(s);
        for(i=1;i<len;i++)
        {
            if(s[i]=='M')
            {
                flag=0;
                break;
            }
            if(s[i]=='I')   cnt++;
            else cnt+=3;
        }
        if(((cnt%6==0)||cnt%2==1&&cnt!=1)||!flag)
        {
            printf("No\n");
        }
        else
            printf("Yes\n");
    }
}