1006 ACM again- Ancient China Mathematics again //The ACM/ICPC Asia Harbin Online First Round Contest 2010 Warmup-2

 

 

ACM again- Ancient China Mathematics again

TimeLimit: 1 Second   MemoryLimit: 32 Megabyte

Totalsubmit: 341   Accepted: 25  

Description

ACM again- Ancient China Mathematics again
ACM again- Ancient China Mathematics again


Dr.K has published a essay on ACM(Ancient China Mathematics), he asked me to thank you for your help what really helped a lot. But another problem raised that the ACM is not as simple as we descripted last time.

In order to help you understand it, we should first go over the ancient people's counting system again:

1、They use only 7 digits to represent numbers, and numbers are as follow:

| -> 1

|| -> 2

||| -> 3

|||| -> 4

||||| -> 5

|||||| -> 6

||||||| -> 7

It is a pity that they have not developed "0" before these people disapeared.

2、If a number is greater than 7, they use digit place just as we do today. Between each two digit places, there is a ",".

eg:

|||||,|| -> 42 (5x8+2)

|||,||,|||| -> 212 (3*64+2*8+4)

In Dr.K's further study, he noticed that, there comes another two symbols - "[" and "]". They are often used together and there is always an expression between them. To see it more clearly, we define it as a grammer expression:

expression = [exp1]

exp1 = [exp1] | [exp1]exp1 | exp1[exp1] | Digit

Digit = "|" | "||" | "|||" | "||||" | "|||||" | "||||||" | "|||||||"

For example:

[||,||||]

[[||][||,|]]

[||[|||][|||||,|]]

are all acceptable expressions.


But what does these expressions mean? That really confused Dr.K a long time, fortunately, he conquered it. He found that:

1、"[]" is used in pairs and all the time with an expersion between them.

2、An expersion is always a number (certainly in their "|||..." form) or another expression.

3、An "exp1[exp2]" or "[exp1]exp2" expersion means exp1 + exp2.

4、An "[exp1][exp2]" expersion means exp1 x exp2.

For example:

[||[||]] = 2+2 = 4

[|] = 1

[[||][|,|]] = 2x9 = 18

[||[[||]|][||]|,||] = 2+(2+1)x2+10 = 18

As same as mordern mathematics, multiplication is piror to addition, and expersion in a "[]" should be count piror to the outer expersion.

For another time, he turns to you for help, it is sure that you will not let him down.

Input

The first line of standard input contains an integer N, the number of test cases. N lines followed.

In each line a squence is given, it is guaranteed that the length of the squence will not exceed 1024 characters and the outcome number will not be greater than 1000000.

Output

For each case you are to output a line giving the number in today's decimal counting system.

Sample Input

3

[||[||]]

[|]

[[||][|,|]]

Sample Output

4

1

18

 

 

 

先ORZ秦牛

#include<stdio.h>

#include<string.h>

char  t[1026];

int stack[1026];

bool sc[1026];

int p;

int main()

{

    int T;

    scanf("%d",&T);

    while(T--)

    {

        p=0;

        memset(sc,0,sizeof(sc));

        scanf("%s",t);

        int l=strlen(t);

        for(int i=0;i<l;i++)

        {

            char a=t[i];

            switch (a)

            {

                case '[':stack[p++]=-1;break;

                case '|':

                {

                    int num=0;

                    while((a=='|'||a==',')&&i<l)

                    {

                        if(a=='|') num++;

                        else num=num<<3;

                        i++;

                        a=t[i];

                    }

                    i--;

                    stack[p++]=num;

                    break;

                }

                case ']':

                {

                    int num=0;

                    while(p>0&&stack[p-1]!=-1)

                    {

                        num+=stack[p-1];

                        sc[p-1]=false;

                        p--;

                    }

                    p--;

                    stack[p]=num;

                    sc[p++]=true;

                    if(p-2>=0&&stack[p-2]!=-1&&sc[p-2])

                    {

                        p--;

                        stack[p-1]*=stack[p];

                        sc[p-1]=true;

                        sc[p]=false;

                    }

                    break;

                }

            }

        }

        int num=0;

        while(p>0) num+=stack[--p];

        printf("%d/n",num);

    }

    return 0;

}