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

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

我要坚持一年，一年后的成功才是我想要的。