算术表达式的求值问题很常见，现在根据课题要求输入一串四则运算表达式，由

The problem of evaluating arithmetic expressions is very common. Now input a string of four arithmetic expressions according to the requirements of the subject, and the computer will find the result of the expression. The input arithmetic expression is required to input a grammatically correct integer arithmetic expression without variable operands in the form of a sequence of characters. In computers, arithmetic expressions consist of constants, variables, operators, and parentheses. Since different operators have different precedences and parentheses must be considered, the evaluation of arithmetic expressions cannot be performed strictly from left to right. Therefore, when programming, it is realized with the help of the stack. At this time, the experiment converts the infix expression into a postfix expression and then outputs it, and then continues the output of the postfix expression. This process is to initialize two stacks: the operator stack s1 and the stack s2 for storing intermediate results; from the left Scan the infix expression to the right; when it encounters the operand, press it into s2; when it encounters the operator, compare it with the priority of the operator on the top of the s1 stack: if s1 is empty, or the operator on the top of the stack is a left parenthesis "(", directly push this operator onto the stack; otherwise, if the priority is higher than the operator on the top of the stack, the operator will also be pushed into s1; otherwise, the operator on the top of the s1 stack will be popped and pushed into s2 , Go to (4.1) again to compare with the new top operator in s1; when encountering parentheses: if it is a left parenthesis "(", then press s1 directly; if it is a right parenthesis ")", then pop s1 in turn The operator at the top of the stack, and push s2 until it encounters the left parenthesis, then discard the pair of parentheses; repeat steps 2 to 5 until the rightmost of the expression; pop the remaining operators in s1 one by one and Push s2; pop the elements in s2 in turn and output. The result is the inverse of the suffix expression corresponding to the infix expression. Continue scanning the suffix expression: if it is a number, let it be pushed onto the stack; if it is an operator, then Take out two operands from the stack, take the first operand as the right operand, and then take the left operand as the left operand, then perform the operation of the operator and push the result into the stack. Repeat the above process until the expression scan is completed. Finally. Only one element left in the stack is the result.

The evaluation of arithmetic expression is very common, now according to the requirements of the subject, input a series of four operation expressions, and get the operation result of the expression by computer. The input arithmetic expression is required to input the integer arithmetic expression with correct syntax and no variable operands in the form of character sequence. In computer, arithmetic expression consists of constant, variable, operator and bracket. Because different operators have different priorities and brackets are considered, it is impossible to evaluate arithmetic expressions strictly from left to right. Therefore, in programming, it is realized by means of stack. In this case, the experiment is to convert infix expression into suffix expression and then output, and then continue to output suffix expression. This process is to initialize two stacks: operator stack S1 and stack S2 storing intermediate results; Scan infix expression from left to right; When the operands are encountered, they are pressed to S2; When encountering an operator, compare its priority with the top of stack operator of S1: if S1 is empty, or the top of stack operator is left bracket "("), put the operator on the stack directly; Otherwise, if the priority is higher than that of the top of stack operator, the operator will also be pushed into S1; Otherwise, the operator at the top of S1 stack will pop up and press into S2, and then go to (4.1) again to compare with the new operator at the top of S1 stack; When encountering brackets: if it is left bracket "("), press S1 directly; If it is "right bracket") ", pop up the operator at the top of S1 stack in turn, and press S2 until the left bracket is met. At this time, the pair of brackets will be discarded; Repeat steps 2 to 5 until the rightmost side of the expression; Pop up the remaining operators in S1 and press them into S2; Pop up the elements in S2 in turn and output. The result inverse is the suffix expression corresponding to infix expression. Continue to scan the suffix expression: if it is a number, let it stack; If it is an operator, two operands are taken from the stack. The first one is taken as the right operand, and the second one is taken as the left operand. Then the operation of the operator is carried out and the result is put on the stack. Repeat the above process until the expression scan is complete. In the end, only one element is left in the stack, which is the result.<br>

The evaluation of arithmetic expressions is very common. Now, according to the requirements of the subject, a series of four arithmetic expressions are input, and the results of the expressions are obtained by the computer. The input arithmetic expression is required to input an integer arithmetic expression with correct syntax and no variables in the form of a sequence of characters. In computers, arithmetic expressions consist of constants, variables, operators and parentheses. Because different operators have different priorities, and parentheses should be considered, the evaluation of arithmetic expressions cannot be strictly carried out from left to right. Therefore, in programming, it is realized by stack. At this time, the experiment is to convert infix expression into suffix expression and then output it, and then continue to output suffix expression. This process is to initialize two stacks: operator stack s1 and stack s2 for storing intermediate results; Scan infix expressions from left to right; When an operand is encountered, it is pressed s2; When an operator is encountered, compare its priority with s1 stack top operator: if s1 is empty, or the stack top operator is left parenthesis "("), this operator will be directly put on the stack; Otherwise, if the priority is higher than that of the stack top operator, press the operator into s1; Otherwise, pop up the top operator of s1 and press it into s2, and go to (4.1) again to compare with the new top operator of s1; When parenthesis is encountered: if it is the left parenthesis "(",press S1 directly; If it is a right parenthesis ")", the operators on the top of s1 stack will pop up in turn, and s2 will be pressed in until the left parenthesis is encountered, at which time the pair of parentheses will be discarded; Repeat steps 2 to 5 until the rightmost part of the expression; Pop up the remaining operators in s1 in turn and press them into S2; The elements in s2 are popped up in turn and output. The result inverse is the suffix expression corresponding to the infix expression. Continue to scan suffix expression: if it is a number, let it stack; If it is an operator, two operands are taken out of the stack, the first one is taken as the right operand, and the second one is taken as the left operand, then the operator is operated and the result is put on the stack. Repeat the above process until the expression scan is completed. There is only one element left in the final stack, which is the result.