C Halting Problem
In computability theory, the halting problem is the problem of determining, from a description of an arbitrary computer program, whether the program will finish running (i.e., halt) or continue to run forever.
Alan Turing proved in 1936 that a general algorithm to solve the halting problem cannot exist, but DreamGrid, our beloved algorithm scientist, declares that he has just found a solution to the halting problem in a specific programming language -- the Dream Language!
Dream Language is a programming language consists of only 5 types of instructions. All these instructions will read from or write to a 8-bit register r, whose value is initially set to 0. We now present the 5 types of instructions in the following table. Note that we denote the current instruction as the i-th instruction.
| Instruction | Description |
|---|---|
| add v | Add v to the register r. As r is a 8-bit register, this instruction actually calculates (r+v)mod256 and stores the result into r, i.e. r←(r+v)mod256. After that, go on to the (i+1)-th instruction. |
| beq v k | If the value of r is equal to v, jump to the k-th instruction, otherwise go on to the (i+1)-th instruction. |
| bne v k | If the value of r isn't equal to v, jump to the k-th instruction, otherwise go on to the (i+1)-th instruction. |
| blt v k | If the value of r is strictly smaller than v, jump to the k-th instruction, otherwise go on to the (i+1)-th instruction. |
| bgt v k | If the value of r is strictly larger than v, jump to the k-th instruction, otherwise go on to the (i+1)-th instruction. |
A Dream Language program consisting of n instructions will always start executing from the 1st instruction, and will only halt (that is to say, stop executing) when the program tries to go on to the (n+1)-th instruction.
As DreamGrid's assistant, in order to help him win the Turing Award, you are asked to write a program to determine whether a given Dream Language program will eventually halt or not.
Input
There are multiple test cases. The first line of the input is an integer T, indicating the number of test cases. For each test case:
The first line contains an integer n (1≤n≤104), indicating the number of instructions in the following Dream Language program.
For the following n lines, the i-th line first contains a string s (s∈{“add”,“beq”,“bne”,“blt”,“bgt”}), indicating the type of the i-th instruction of the program.
- If s equals to "add", an integer v follows (0≤v≤255), indicating the value added to the register;
- Otherwise, two integers v and k follow (0≤v≤255, 1≤k≤n), indicating the condition value and the destination of the jump.
It's guaranteed that the sum of n of all test cases will not exceed 105.
Output
For each test case output one line. If the program will eventually halt, output "Yes" (without quotes); If the program will continue to run forever, output "No" (without quotes).
Sample Input
4
2
add 1
blt 5 1
3
add 252
add 1
bgt 252 2
2
add 2
bne 7 1
3
add 1
bne 252 1
beq 252 1
Sample Output
Yes
Yes
No
No
Hint
For the second sample test case, note that r is a 8-bit register, so after four "add 1" instructions the value of r will change from 252 to 0, and the program will halt.
For the third sample test case, it's easy to discover that the value of r will always be even, so it's impossible for the value of r to be equal to 7, and the program will run forever.
一道纯模拟题目。假设你手上有数r，r初始为0，题目给出了五个操作，
add x ：r+=x（并且要模256）
deq v x 如果v==r就 跳到第x个操作
········
题目的麻烦之处在于如何判定失败，我们知道如果陷入无限循环也是失败。那么我们就需要记录每一层的操作，然后再在进行这次操作的时候，查看是否之前有过相同状态也就是手里的 r 与之前的一次r相等且是同一层，就可以确定陷入循环，直接结束输出 No就行。
#include<bits/stdc++.h>
using namespace std;
struct node{
string a;
int x,y;
int r[300];
};
node s[10005];
int main()
{
int t;
cin>>t;
while(t--)
{
string a;
int n;
cin>>n;
for(int i=1;i<=n;++i)
{
cin>>s[i].a;
if(s[i].a=="add"){
cin>>s[i].x;
}else if(s[i].a=="beq")
{
cin>>s[i].x;
cin>>s[i].y;
for(int j=0;j<258;j++) s[i].r[j]=0;
}else if(s[i].a=="bne")
{
cin>>s[i].x;
cin>>s[i].y;
for(int j=0;j<258;j++) s[i].r[j]=0;
}else if(s[i].a=="blt")
{
cin>>s[i].x;
cin>>s[i].y;
for(int j=0;j<258;j++) s[i].r[j]=0;
}else if(s[i].a=="bgt")
{
cin>>s[i].x;
cin>>s[i].y;
for(int j=0;j<258;j++) s[i].r[j]=0;
}
}
int k=0,r=0,i=1;
while(1)
{
if(i>=n+1)
{
cout<<"Yes\n";
break;
}
if(s[i].a=="add"){
r+=s[i].x;
r%=256;
++i;
}else if(s[i].a=="beq")
{
if( s[i].r[r]==1) {
cout<<"No\n";
break;
}
s[i].r[r]=1;
if(r==s[i].x)
{
i=s[i].y;
}else {
++i;
}
}else if(s[i].a=="bne")
{
if( s[i].r[r]==1) {
cout<<"No\n";
break;
}
s[i].r[r]=1;
if(r!=s[i].x)
{
i=s[i].y;
}else {
++i;
}
}else if(s[i].a=="blt")
{
if( s[i].r[r]==1) {
cout<<"No\n";
break;
}
s[i].r[r]=1;
if(r<s[i].x)
{
i=s[i].y;
}else {
++i;
}
}else if(s[i].a=="bgt")
{
if( s[i].r[r]==1) {
cout<<"No\n";
break;
}
s[i].r[r]=1;
if(r>s[i].x)
{
i=s[i].y;
}else {
++i;
}
}
}
}
}