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

 1 #include<cstdio>

 2 #include<iostream>

 3 #include<algorithm>

 4 #include<cstring>

 5 #include<cmath>

 6 #include<queue>

 7 #include<map>

 8 using namespace std;

 9 #define MOD 1000000007

10 const int INF=0x3f3f3f3f;

11 const double eps=1e-5;

12 typedef long long ll;

13 #define cl(a) memset(a,0,sizeof(a))

14 #define ts printf("*****\n");

15 const int MAXN=1005055;

16 int n,m,tt;

17 char s[MAXN];

18 int main()

19 {

20     int i,j,k;

21     #ifndef ONLINE_JUDGE

22     freopen("1.in","r",stdin);

23     #endif

24     scanf("%d",&tt);

25     while(tt--)

26     {

27         scanf("%s",s);

28         if(s[0]!='M')

29         {

30             printf("No\n");

31             continue;

32         }

33         int cnt=0;

34         bool flag=1;

35         int len=strlen(s);

36         for(i=1;i<len;i++)

37         {

38             if(s[i]=='M')

39             {

40                 flag=0;

41                 break;

42             }

43             if(s[i]=='I')   cnt++;

44             else cnt+=3;

45         }

46         if(((cnt%6==0)||cnt%2==1&&cnt!=1)||!flag)

47         {

48             printf("No\n");

49         }

50         else

51             printf("Yes\n");

52     }

53 }