题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4662
MU Puzzle
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1296 Accepted Submission(s): 539
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
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规律题，把u所有化成i，然后多写下，就能发现规律
#include<iostream>
#include<cstring>
using namespace std;
char str[1000010];
int main()
{
int n;
cin>>n;
while(n--)
{
cin>>str;
int l=strlen(str),xi=0,xu=0,k=0;
if(str[0]!='M')
{
cout<<"No"<<endl;
continue;
}
for(int i=1;i<l;i++)
{
if(str[i]=='I')
xi++;
else if(str[i]=='U')
xu++;
else
{
k=1;
break;
}
}
if(k==1 || ((xi+xu)%2==1 && !(xi==1 && xu==0)) || xi%3==0)
{
cout<<"No"<<endl;
continue;
}
else
{
cout<<"Yes"<<endl;
continue;
}
}
return 0;
}