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Computer Networking: A Top-Down Approach (7th Edition)
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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In this challenge, you will evaluate an arithmetic expression. However the given expression may not be well formed, and you also want to check if this happens. If the expression has mismatched brackets, or it is missing operator or operand, then it is invalid. Formally, an expression E is defined as: E - E + E | E – E | E * E | (E) | integer Note that E must be non-empty, and a pair of brackets must not contain empty content. Standard input The input has a single integer T on the first line, the number of expressions to evaluate. Each expression to evaluate is given on a single line. An expression is a non-empty string without spaces. It consists of digits giving integer operands, round brackets, and arithmetic operators +, -, and *. Standard output For each expression, if it is valid, output the result of its evaluation. Since the result may be very large, output the result modulo 1 000 000 007 (10° + 7). If the expression is invalid, output invalid. Constraints and notes • 1<T< 25 • The length of an expression does not exceed 10°. • All the integers in the expression are smaller than 1 000 and do not have leading zeroes, even when the expression is invalid. • Note that all arithmetic operators are binary. That is, there are no unary negations and -3, -1*2, 1--2 are invalid. • In modular arithmetic the result is always non-negative. If a - b< 0(0 < a, b < P), then the modular result (a b) mod P = (P+a – b) mod P. • For 20% of the test files, there are no brackets in the given expression.
Transcribed Image Text:In this challenge, you will evaluate an arithmetic expression. However the given expression may not be well formed, and you also want to check if this happens. If the expression has mismatched brackets, or it is missing operator or operand, then it is invalid. Formally, an expression E is defined as: E - E + E | E – E | E * E | (E) | integer Note that E must be non-empty, and a pair of brackets must not contain empty content. Standard input The input has a single integer T on the first line, the number of expressions to evaluate. Each expression to evaluate is given on a single line. An expression is a non-empty string without spaces. It consists of digits giving integer operands, round brackets, and arithmetic operators +, -, and *. Standard output For each expression, if it is valid, output the result of its evaluation. Since the result may be very large, output the result modulo 1 000 000 007 (10° + 7). If the expression is invalid, output invalid. Constraints and notes • 1<T< 25 • The length of an expression does not exceed 10°. • All the integers in the expression are smaller than 1 000 and do not have leading zeroes, even when the expression is invalid. • Note that all arithmetic operators are binary. That is, there are no unary negations and -3, -1*2, 1--2 are invalid. • In modular arithmetic the result is always non-negative. If a - b< 0(0 < a, b < P), then the modular result (a b) mod P = (P+a – b) mod P. • For 20% of the test files, there are no brackets in the given expression.
Input Output Explanation 4 11 There are 4 test cases without round 1+2*3+4 1000000006 brackets. 1*2-3 123 • Case 1: It is straightforward to evaluate that 1+2 x 3 +4 = 123 invalid +-* 11. • Case 2:1 x 2 – 3 = -1. In modular arithmetic, -1 is 1 000 000 006 (mod 1 000 000 007). • Case 3: By definition E → integer. Therefore 123 evaluates to 123. • Case 4: Each operator must have operands. This expression is invalid. 4 9 There are 4 test cases with round (1+2)*3 invalid brackets. () invalid () 100 (((1))) • Case 1: It is straightforward to evaluate that (1+ 2) × 3 = 9. • Case 2: Empty brackets are not 1 allowed by definition. • Case 3: Not only empty brackets are disallowed, but also there misses an operator between () and 100. • Case 4: By definition E (E). Therefore E → (E) → ((E)) → (((E))) → (((1))), which evaluates to 1.
Transcribed Image Text:Input Output Explanation 4 11 There are 4 test cases without round 1+2*3+4 1000000006 brackets. 1*2-3 123 • Case 1: It is straightforward to evaluate that 1+2 x 3 +4 = 123 invalid +-* 11. • Case 2:1 x 2 – 3 = -1. In modular arithmetic, -1 is 1 000 000 006 (mod 1 000 000 007). • Case 3: By definition E → integer. Therefore 123 evaluates to 123. • Case 4: Each operator must have operands. This expression is invalid. 4 9 There are 4 test cases with round (1+2)*3 invalid brackets. () invalid () 100 (((1))) • Case 1: It is straightforward to evaluate that (1+ 2) × 3 = 9. • Case 2: Empty brackets are not 1 allowed by definition. • Case 3: Not only empty brackets are disallowed, but also there misses an operator between () and 100. • Case 4: By definition E (E). Therefore E → (E) → ((E)) → (((E))) → (((1))), which evaluates to 1.
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