MU Puzzle
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 299 Accepted Submission(s): 157
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.
Total length of all strings <= 106.
Output
n lines, each line is 'Yes' or 'No'.
Sample Input
2
MI
MU
Sample Output
Yes
No
Source
多校第六场也结束了，我们队每场都只能过一两道题，感觉前途渺茫啊……
第六场的1008，这场比赛我唯一AC的一道题，也是我们队唯一AC的一道题
给一个字符串，问它能否由MI按照题干中给的规则转换过来
我的做法是，由规则逆向思考，先所给串中所有的U全部换成3个I，然后统计I的个数。
由第一条规则和第三条规则易知I的个数一定能写成（2^x-6*y）的形式，否则不能由MI转化过来
因此只要判断I的个数和比刚好比它大的两个2^x的差能否被六整除即可。（注意这里一定要判断2个2^x，因为2^x对六的模可能是2或4，交替出现）
1 #include<iostream> 2 #include<cstring> 3 #include<cstdio> 4 5 using namespace std; 6 7 int len; 8 int two[50]; 9 char s[1000100]; 10 11 void initial() 12 { 13 two[0]=1; 14 for(int i=1;i<=30;i++) 15 two[i]=two[i-1]*2; 16 } 17 18 int main() 19 { 20 int t; 21 22 initial(); 23 24 scanf("%d",&t); 25 getchar(); 26 27 while(t--) 28 { 29 gets(s); 30 if(s[0]!='M') 31 { 32 cout<<"No"<<endl; 33 continue; 34 } 35 int len,ans=0; 36 bool flag=true; 37 len=strlen(s); 38 for(int i=1;i<len&&flag;i++) 39 if(s[i]=='I') 40 ans++; 41 else if(s[i]=='U') 42 ans+=3; 43 else if(s[i]=='M') 44 { 45 cout<<"No"<<endl; 46 flag=false; 47 } 48 if(!flag) 49 continue; 50 int i; 51 for(i=0;i<=30;i++) 52 if(two[i]>=ans) 53 break; 54 if((two[i]-ans)%6==0||(two[i+1]-ans)%6==0) 55 cout<<"Yes"<<endl; 56 else 57 cout<<"No"<<endl; 58 } 59 60 return 0; 61 }