HDU 4662 MU Puzzle

MU Puzzle

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 81    Accepted Submission(s): 44

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

2MIMU

 

Sample Output

YesNo

 

Source

2013 Multi-University Training Contest 6

 

Recommend

zhuyuanchen520

 

题意：  

1、Mx -> Mxx

2、III -> U

3、UU -> 空

问MI 能否进过上面操作转化成  读入的字符串。

思路：模拟

根据3个操作可知：

1、M前面的不能有字母。

2、只能有一个M

3、操作一只能得到 2^n个I (n > 0)

4、可以消去6个I.

然后得到一个公式  2^n - 6 * m = sum; (sum 表示读入的字符串中I的个数);

#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>#include <vector>#include <cmath>using namespace std;//const int V = 1000000 + 50;const int MaxN = 80 + 5;const int mod = 10000 + 7;const __int64 INF = 0x7FFFFFFFFFFFFFFFLL;const int inf = 0x7fffffff;int n, flag;char ch[V];int main() {    int i, j;    scanf("%d", &n);    while(n--) {        flag = 1;        scanf("%s", &ch);        if(ch[0] != 'M')            printf("No\n");        else{            int sum = 0;            for(i = 1; ch[i]; ++i) {                if(ch[i] == 'M') {                    flag = 0;                    break;                }                else if(ch[i] == 'U')                    sum += 3;                else                    sum ++;            }            if(!flag) {                printf("No\n");                continue;            }            if(sum == 1) {                printf("Yes\n");                continue;            }            if(sum % 2 == 1) {                printf("No\n");                continue;            }            for(i = 25; i >= 0; --i) {                if(((1 << i) - sum % 6) == 0) {                    flag = 0;                    break;                }            }            if(flag)                printf("No\n");            else                printf("Yes\n");        }    }}