zoj 4049

Halting Problem

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Time Limit: 1 Second      Memory Limit: 65536 KB

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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 , 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 -th instruction.

Instruction	Description
add	Add  to the register . As  is a 8-bit register, this instruction actually calculates  and stores the result into , i.e. . After that, go on to the -th instruction.
beq	If the value of  is equal to , jump to the -th instruction, otherwise go on to the -th instruction.
bne	If the value of  isn't equal to , jump to the -th instruction, otherwise go on to the -th instruction.
blt	If the value of  is strictly smaller than , jump to the -th instruction, otherwise go on to the -th instruction.
bgt	If the value of  is strictly larger than , jump to the -th instruction, otherwise go on to the -th instruction.

A Dream Language program consisting of  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 -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 , indicating the number of test cases. For each test case:

The first line contains an integer  (), indicating the number of instructions in the following Dream Language program.

For the following  lines, the -th line first contains a string  (), indicating the type of the -th instruction of the program.

• If  equals to "add", an integer  follows (), indicating the value added to the register;

• Otherwise, two integers  and  follow (, ), indicating the condition value and the destination of the jump.

It's guaranteed that the sum of  of all test cases will not exceed .

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  is a 8-bit register, so after four "add 1" instructions the value of  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  will always be even, so it's impossible for the value of  to be equal to 7, and the program will run forever.

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Author: WENG, Caizhi
Source: The 2018 ACM-ICPC Asia Qingdao Regional Contest, Online

 

const int N =1e4+5; 
int vis[N][260];
int t,n;
struct Ma{
    char s[10];
    int v,k;
}ma[N];
void  init()
{
    for(int i=1;i<=n;i++)
    {
        for(int j=0;j<260;j++){//一开始写成了26
            vis[i][j]=0;
        }
    }
}
int  main()
{
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        init();
        for(int i=1;i<=n;i++)
        {
            scanf("%s",ma[i].s);
            if(ma[i].s[0]=='a') {
                scanf("%d",&ma[i].v);
                ma[i].k=0;
            }
            else{
                scanf("%d%d",&ma[i].v,&ma[i].k);
            }
        }
        int pos=1,ret=0,flag=0;
        while(1)
        {
            if(pos==n+1) {
                flag=1;
                break;
            }
            if(vis[pos][ret]) {
                flag=0;
                break;
            }
            vis[pos][ret]++;//由ret到第pos个指令，若重复出现就会死循环
            if(ma[pos].s[0]=='a') {
                ret=(ret+ma[pos].v)%256;
                pos++;
            }
            else if(ma[pos].s[1]=='e'){
                if(ret==ma[pos].v){
                    pos=ma[pos].k;
                }
                else{
                    pos++;
                }
            }
            else if(ma[pos].s[1]=='n'){
                if(ret!=ma[pos].v){
                    pos=ma[pos].k;
                }
                else{
                    pos++;
                }
            }
            else if(ma[pos].s[1]=='l'){
                if(ret<ma[pos].v){
                    pos=ma[pos].k;
                }
                else{
                    pos++;
                }
            }
            else if(ma[pos].s[1]=='g'){
                if(ret>ma[pos].v){
                    pos=ma[pos].k;
                }
                else{
                    pos++;
                }
            }
        }
        printf("%s\n",flag==1?"Yes":"No");
    }
    return 0;
}