HDU 4662 MU Puzzle (2013多校6 1008 水题)

MU Puzzle

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 134    Accepted Submission(s): 71

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

2
MI
MU

 

Sample Output

Yes
No

 

Source

2013 Multi-University Training Contest 6

 

Recommend

zhuyuanchen520

 

 

 

 

 

全场最水的题目。

 

 

一开始时MI。

其实可以全部转化成I的个数。

 

一个U相当于3个I，减少两个U，相当于减少6个I。

 

第一个操作相当于可以把I的个数翻倍。

 

 

所以1个I，2个I，4个I，8,16,32，。。。。这些2^n个I都是可以实现的。

 

 

而且可以减掉6个I，8-6=2,16-6=10,10-6=4，等等都可以。

 

10个I可以，20个I也可以。

 

 

这样数I的个数，比赛时候暴力预处理下所以可以的情况，就像晒素数一样。

 

 

 

其实可以发现规律的，除了一个I之外，奇数个I和个数被6整除的都不行，可以根据整除性分析下。

 

 

 

 

 

 

 

贴上代码，里面有一个init函数式预处理的，虽然没有用上

 

 

 

 /*
  * Author:  kuangbin
  * Created Time:  2013/8/8 11:54:08
  * File Name: 1008.cpp
  */
 #include <iostream>
 #include <cstdio>
 #include <cstdlib>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <string>
 #include <vector>
 #include <stack>
 #include <queue>
 #include <set>
 #include <time.h>
 using namespace std;
 char str[];
 const int MAXN = ;
 bool dp[MAXN];
 bool vis[MAXN];
 void init()
 {
     memset(dp,false,sizeof(dp));
     memset(vis,false,sizeof(vis));
     dp[] = true;
     queue<int>q;
     q.push();
     vis[] = true;
     while(!q.empty())
     {
         int tmp = q.front();
         dp[tmp] = true;
         q.pop();
         if(tmp == ){dp[] = true;vis[] = true;continue;}
         if(tmp >=  && !vis[tmp-])
         {
             q.push(tmp-);
             vis[tmp-] = true;
         }
         tmp *= ;
         while(tmp < MAXN)
         {
             if(vis[tmp])break;
             dp[tmp] = true;
             if(tmp >=  && !vis[tmp-])
             {
                 q.push(tmp-);
                 vis[tmp-] = true;
             }
             tmp *= ;
         }
     }
 }

 int main()
 {
     int T;
     init();
     scanf("%d",&T);
     while(T--)
     {
         bool flag = true;
         scanf("%s",str);
         int len = strlen(str);
         if(str[]!='M')
         {
             printf("No\n");
             continue;
         }
         int cnt = ;
         for(int i = ;i < len;i++)
         {
             if(str[i]=='M')
             {
                 flag = false;
                 break;
             }
             if(str[i] =='I')cnt++;
             else cnt+= ;
         }
         if(!flag)
         {
             printf("No\n");
             continue;
         }
         if(cnt == )
         {
             printf("Yes\n");
             continue;
         }
         if(cnt% ==  || cnt% == )printf("No\n");
         else printf("Yes\n");
     }
     return ;
 }