Description

Given an infix Boolean expression with parentheses, which has at most 4 variables ‘A’, ’B’, ‘C’, and ‘D’, and two operators ‘&’ and ‘|’. Build a corresponding syntax tree for it and print it's truth table.

For example, if input is "(A&C)|(A|B)", then result will be

A B C D output

0 0 0 0 0
0 0 0 1 0
0 0 1 0 0
0 0 1 1 0
0 1 0 0 1
0 1 0 1 1
0 1 1 0 1
0 1 1 1 1
1 0 0 0 1
1 0 0 1 1
1 0 1 0 1
1 0 1 1 1
1 1 0 0 1
1 1 0 1 1
1 1 1 0 1
1 1 1 1 1

 

To parse the infix expression with parentheses, use the grammar below.

EXPR = FACTOR | EXPR OP FACTOR

FACTOR = ID | (EXPR)

EXPR is the expression, ID is one of ‘A’, ‘B’, ’C’, or ‘D’, and OP is one of ‘&’ or ‘|’.

 

You will be provided with main.c and function.h. main.c contains the implementation of functions printPrefix and freeTree. function.h contains the definition of tree node and variables. You only need to implement the following four functions in function.c.

BTNode* makeNode(char c)   // create a node without any child

BTNode* EXPR()   // parse an infix expression and generate a syntax tree

BTNode* FACTOR()   // use the parse grammar FACTOR = ID | (EXPR) to deal with parentheses

void printResult(BTNode *root)    // print the truth table

 

For OJ submission:

        Step 1. Submit only your function.c into the submission block.(Please choose C compiler)

        Step 2. Check the results and debug your program if necessary.

main.c

#include <stdio.h> #include <string.h> #include <stdlib.h> #include "function.h" int main() { sym[0]='A'; sym[1]='B'; sym[2]='C'; sym[3]='D'; sym[4]='&'; sym[5]='|'; while (scanf("%s", expr)!=EOF) { pos = strlen(expr) - 1; BTNode *root = EXPR(); printResult(root); printf("\n"); freeTree(root); } return 0; } /* clean a tree.*/ void freeTree(BTNode *root) { if (root!=NULL) { freeTree(root->left); freeTree(root->right); free(root); } }

function.h

#include <stdio.h> #include <stdlib.h> #define MAXEXPR 256 #define NUMSYM 6 char expr[MAXEXPR]; // string to store the input expression. int pos; // current position of parsing, from end to start typedef enum {ID_A, ID_B, ID_C, ID_D, OP_AND, OP_OR} TokenSet; char sym[NUMSYM]; typedef struct _Node { TokenSet data; struct _Node *left, *right; } BTNode; BTNode* EXPR(); BTNode* FACTOR(); BTNode* makeNode(char c); void freeTree(BTNode *root); void printResult(BTNode *root);

Input

The input contains a sequences of prefix expression. It only has at most 4 variables ‘A’, ‘B’, ‘C’, and ‘D’, and 2 operators, AND ‘&’ and OR ‘|’. The length of prefix expression is shorter than 30.

Output

It has 5 variables 'A', 'B', 'C', 'D', and output, each variable is separated by a space.

Sample Input  Download

Sample Output  Download

Partial Judge Code

10982.c

Partial Judge Header

10982.h

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Description

Linked list

typedef struct _Node {
    int data;
    struct _Node *next;
} Node;

 

/* print the content of the linked list.*/

void printList(Node *head)
{
    Node *temp;
    for(temp=head; temp!=NULL; temp=temp->next) {
        printf("%d ", temp->data);
    }
}

/*Insert a node after the node pointed by nd.

Return 1 if insertion succeeds.
Otherwise, return 0.*/

int insertNode(Node* nd, int data)
{
    if (nd != NULL) {

        Node* temp = (Node*)malloc(sizeof(Node));
        temp->data = data;
        temp->next = nd->next;
        nd->next = temp;
        return 1;
    }
    return 0;
}

 

/*Delete a node after the node pointed by nd.

Return 1 if insertion succeeds. 

Otherwise, return 0.*/

int deleteNode(Node* nd)
{
    if (nd != NULL) {
        Node* temp = nd->next;
        if (temp != NULL) {
            nd -> next = temp->next;
            free(temp);
            return 1;
        }
    }
    return 0;
}

 

Binary tree

typedef struct _node {
     int data;
     struct _node *left, *right;
} BTNode;

 

Tree traversal

• Pre-order: visit root, left subtree, and right subtree
• In-order: visit left subtree, root, and right subtree
• Post-order: visit left subtree, right subtree, and root

 

Expression order

• Prefix:  operator, left operand,and right operand
• Infix:  left operand, operator, and right operand
• Postfix: left operand, right operand, and operator

 

Syntax tree

The internal nodes are operators, and the leaves are operands.

Recursive evaluation of prefix expression

/*It is a pseudo code */

int evalBoolExpr(){

char c = getchar();

If c is an operator

   op1 = evalBoolExpr();

   op2 = evalBoolExpr();

   return op1 c op2;

Else if c is a variable

   return the value of c;

}

Grammar of infix expression with parenthesis

• EXPR = FACTOR| EXPR OP FACTOR
• FACTROR = ID | (EXPR)

 

/*Suppose expr[] is an array of expression and pos is the current position, which initially is strlen(expr)-1. */

/*-------------------------------------*/

BTNode* makeNode(char c)
{
    int i;
    BTNode *node = (BTNode*)malloc(sizeof(BTNode));
    set node->data
    node->left = NULL;
    node->right = NULL;
    return node;
}

 

/*-------------------------------------*/

BTNode* FACTOR()
{
    char c;
    BTNode *node = NULL;
    if (pos>=0) {
        c = expr[pos--];
        if (c is an ID) {
            node = makeNode(c);
        } else if (c==‘)’) {
            node = EXPR();
            if(expr[pos--]!= '(') {
                printf("Error!\n");
                freeTree(node);
            }
        }
    }
    return node;
}

 

/*-------------------------------------*/

BTNode* EXPR()
{

    char c;

    BTNode *node = NULL, *right=NULL;

    if (pos>=0) {

        node = right = FACTOR();

        if (pos>0) {

            c = expr[pos];

            if (c is an operator) {

                node = makeNode(c);

                node->right = right;

                pos--;

                node->left = EXPR();

            }

        }

    }

    return node;

}

Input

Output

Sample Input  Download

Sample Output  Download

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Description

Please create a binary search tree, and support insert operation. Please ignore the repeat number and print out the tree with preorder sequence. (The root node is greater than the node in the left subtree and smaller than the node in the right subtree.)

Pre-order:

1. Check if the current node is empty / null.
2. Display the data part of the root (or current node).
3. Traverse the left subtree by recursively calling the pre-order function.
4. Traverse the right subtree by recursively calling the pre-order function

Input

The input contains two lines. The first line is a integer n (n<=8000) , and the second line is a sequence S, where S contain n non-negative integer number. (The first number in the sequence S is the root.) 

Output

Please print out the tree with preorder. (You don't need to write the print_tree(). You only need to create a binary search tree.)

Sample Input  Download

Sample Output  Download

Partial Judge Code

11372.c

Partial Judge Header

11372.h

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Description

You are required to use linked list to do the addition of two polynomials.

Input

The input contains two lines. Each lines presents a polynomial. The format of each line is looked like : "5 4 -3 2 1 0" which means polynomial "5x⁴-3x²+1x⁰".

Each polynomial must contain a constant term. (You can use this rule to determine whether the polynomial is end.)

For example, "-2 3 1 1 0 0" should be -2x³+1x¹.

(The input polynomial should be arrangement in descending power.)

Output

Output the answer. Print a space character (' ') in the begining.

For example, if the input is

5 4 -3 2 -1 1 1 0 (means 5x⁴-3x²-1x¹+1)

-2 3 1 1 0 0 (means -2x³+1x¹)

the output should be 

" 5 4 -2 3 -3 2 1 0" (which means 5x⁴-2x³-3x²+1).

If the value of coefficient is 0, you don't have to print it.

(The output polynomial should be arrangement in descending power.)

Sample Input  Download

Sample Output  Download

Partial Judge Code

11373.c

Partial Judge Header

11373.h

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