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?
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.
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
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
1 /* 2 * Author: kuangbin 3 * Created Time: 2013/8/8 11:54:08 4 * File Name: 1008.cpp 5 */ 6 #include <iostream> 7 #include <cstdio> 8 #include <cstdlib> 9 #include <cstring> 10 #include <cmath> 11 #include <algorithm> 12 #include <string> 13 #include <vector> 14 #include <stack> 15 #include <queue> 16 #include <set> 17 #include <time.h> 18 using namespace std; 19 char str[1000010]; 20 const int MAXN = 3000010; 21 bool dp[MAXN]; 22 bool vis[MAXN]; 23 void init() 24 { 25 memset(dp,false,sizeof(dp)); 26 memset(vis,false,sizeof(vis)); 27 dp[1] = true; 28 queue<int>q; 29 q.push(1); 30 vis[1] = true; 31 while(!q.empty()) 32 { 33 int tmp = q.front(); 34 dp[tmp] = true; 35 q.pop(); 36 if(tmp == 0){dp[0] = true;vis[0] = true;continue;} 37 if(tmp >= 6 && !vis[tmp-6]) 38 { 39 q.push(tmp-6); 40 vis[tmp-6] = true; 41 } 42 tmp *= 2; 43 while(tmp < MAXN) 44 { 45 if(vis[tmp])break; 46 dp[tmp] = true; 47 if(tmp >= 6 && !vis[tmp-6]) 48 { 49 q.push(tmp-6); 50 vis[tmp-6] = true; 51 } 52 tmp *= 2; 53 } 54 } 55 } 56 57 58 int main() 59 { 60 int T; 61 init(); 62 scanf("%d",&T); 63 while(T--) 64 { 65 bool flag = true; 66 scanf("%s",str); 67 int len = strlen(str); 68 if(str[0]!='M') 69 { 70 printf("No "); 71 continue; 72 } 73 int cnt = 0; 74 for(int i = 1;i < len;i++) 75 { 76 if(str[i]=='M') 77 { 78 flag = false; 79 break; 80 } 81 if(str[i] =='I')cnt++; 82 else cnt+= 3; 83 } 84 if(!flag) 85 { 86 printf("No "); 87 continue; 88 } 89 if(cnt == 1) 90 { 91 printf("Yes "); 92 continue; 93 } 94 if(cnt%2 == 1 || cnt%6 == 0)printf("No "); 95 else printf("Yes "); 96 } 97 return 0; 98 }