最后一周第二天训练赛之第二题

试题：

B - B

Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu

Submit  Status  Practice  SPOJ ICODER

Description

Mathews uses a brand new 16-bit instruction processor. (Yeah i am being sarcastic!). It has one register (say R) and it supports two instructions: 

• ADD X; Impact: R = (R + X) mod 65536
• MUL X; Impact: R = (R * X) mod 65536
• [For both instructions 0 <= X <= 65535]

Mathews sees a segment of code, but doesnot know what value the register had before the code was being executed. How many possible values can the register have after the segment completed execution?

 

Input Format:
The input file consists of multiple testcases. 
The first line of each testcase contains one integer,  N. (1 <= N <= 100,000).
The following  N lines contain one instructions each. 
Input terminates with a line containing N=0, which must not be processed. 

Output Format:
For each testcase print one integer in a single line, denoting the number of different values the register can take after code execution.

Sample Input:

1

ADD 3

1

MUL 0

5

MUL 3

ADD 4

MUL 5

ADD 3

MUL 2

8

ADD 32

MUL 5312

ADD 7

MUL 7

ADD 32

MUL 5312

ADD 7

MUL 7

0

Sample Output:

65536

1

32768

16
这道题是一道传说中的水题吧，但是写的时候还是有很大的问题的。没有AC。就当做一个记录留下来吧。
AC代码：

 1 #include <iostream>

 2 #include <cstdio>

 3 

 4 using namespace std;

 5 char ans[6];

 6 int main()

 7 {

 8     int n;

 9     while(cin>>n&&n){

10         int aa = 65536,haha;

11         for(int i = 0;i < n;i++){

12         scanf("%s",ans);

13         cin>>haha;

14         if(ans[0]=='M'){

15 

16             if(haha==0){aa = 1;break;}

17             while(haha%2==0){

18                 haha/=2;

19                 aa/=2;

20             }

21         }

22         }

23         if(aa>=1)

24         cout<<aa<<endl;

25         else cout<<1<<endl;

26     }

27     return 0;

28 }

AC代码