Sicily 1268. Single Digit Adder

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1268. Single Digit Adder

Constraints

Time Limit: 1 secs, Memory Limit: 32 MB

Description

Write a program that can evaluate expressions from the following roughly BNF (Backus Naur Form) grammar:

expr ::= term | expr `+' term | expr `-' termunary_op ::= `+' term | `-' termterm ::= `(' expr `)' | `(' unary_op `)' | literalliteral ::= [0-9]

There will be no whitespace within an expression. All expressions will consist solely of the characters `(', `)', `+', `-', and the digits 0 through 9. You may assume that all input is well-formed.

Input

The input will consist of one expression per line followed by a newline. There will be no blank lines in the file.

Output

For each expression, output its integer value, followed by a single newline.

Sample Input

1 (-2)+3 (1-(2+3)) (1-2+3) (1-(+(2-3)))

Sample Output

1 1 -4 2 2
// Problem#: 1268// Submission#: 3590564// The source code is licensed under Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License// URI: http://creativecommons.org/licenses/by-nc-sa/3.0/// All Copyright reserved by Informatic Lab of Sun Yat-sen University#include <string>#include <stack>#include <iostream>#include <vector>#include <string.h>#include <stdlib.h>#include <stdio.h>using namespace std;vector<string> ans_vector_post;string post_string;inline int prior(char op) {  // 计算优先级函数    if (op == '+' || op == '-') {        return 1;    } else if (op == '*' || op == '/' || op == '%') {        return 2;    } else {        return 0;    }}long long int string_to_int(string in) {  // 将输入的字符串转化为相应数字函数    char s[50];    for (int i = 0; i < 50; i++) {        s[i] = '\0';    }    for (int i = 0; i < in.size(); i++) {        s[i] = in[i];    }    long long int ans;    sscanf(s, "%lld", &ans);    return ans;}string trans_to_postfix_expression_to_s(string in) {        stack<char> op;  // 操作符栈    ans_vector_post.clear();  // 后缀表达式存放数组    for (int i = 0; i < in.size();) {        char c = in[i];        if ('0' <= c && c <= '9') {  // 是数字直接插入            string num;            int j;            for (j = i; j < in.size() && '0' <= in[j] && in[j] <= '9'; j++) {                num.push_back(in[j]);            }            ans_vector_post.push_back(num);            i = j;        } else {            if (c == '(') {  // 是开括号直接插入                op.push('(');            } else {                if (c == ')') {  // 是闭括号就把原本栈中的运算符都输出，直到遇到开括号，注意开括号要丢弃                    while (op.top() != '(') {                        string temp;                        temp.push_back(op.top());                        ans_vector_post.push_back(temp);                        op.pop();                    }                    op.pop();                } else {  // 假如是加减乘除取余                    if (op.empty()) {  // 操作符栈是空就直接插入                        op.push(c);                    } else {  // 如果扫描到的运算符优先级高于栈顶运算符则，把运算符压入栈。否则的话，就依次把栈中运算符弹出加到数组ans的末尾，直到遇到优先级低于扫描到的运算符或栈空，并且把扫描到的运算符压入栈中                        if (prior(c) > prior(op.top())) {                            op.push(c);                        } else {                            while (!op.empty() && prior(c) <= prior(op.top())) {                                string temp;                                temp.push_back(op.top());                                ans_vector_post.push_back(temp);                                op.pop();                            }                            op.push(c);                        }                    }                }            }            i++;        }    }    while (!op.empty()) {  // 注意把操作符栈中的剩余操作符输出        string temp;        temp.push_back(op.top());        ans_vector_post.push_back(temp);        op.pop();    }        post_string.clear();  // 构造string并返回    for (int i = 0; i < ans_vector_post.size(); i++) {        post_string += ans_vector_post[i];    }        return post_string;}long long int calculate_from_postfix_expression() {        //利用栈对后缀表达式求值，直接从后缀表达式的左往右扫描，遇到数字放入栈中，遇到字符就把栈顶的两个数字拿出来算，然后再放进栈        stack<long long int> ans_post;    for (int i = 0; i < ans_vector_post.size(); i++) {        long long int x, y;        if ('0' <= ans_vector_post[i][0] && ans_vector_post[i][0] <= '9') {            ans_post.push(string_to_int(ans_vector_post[i]));        } else {            y = ans_post.top();  // 注意顺序，这里好比xy+就是x+y            ans_post.pop();            x = ans_post.top();            ans_post.pop();            if (ans_vector_post[i][0] == '+') {                ans_post.push(x + y);            } else if (ans_vector_post[i][0] == '-') {                ans_post.push(x - y);            } else if (ans_vector_post[i][0] == '*') {                ans_post.push(x * y);            } else if (ans_vector_post[i][0] == '/') {                ans_post.push(x / y);            } else {                ans_post.push(x % y);            }        }    }    return ans_post.top();}string predo(string in) {    int s = in.size();    string ret;    for (int i = 0; i < s; i++) {        if (i == 0 && (in[i] == '-' || in[i] == '+')) {            ret.push_back('0');            ret.push_back(in[i]);        } else if (i > 0 && (in[i] == '-' || in[i] == '+') && in[i - 1] == '(') {            ret.push_back('0');            ret.push_back(in[i]);        } else ret.push_back(in[i]);    }    return ret;}int main() {    std::ios::sync_with_stdio(false);    while (1) {        string in;        cin >> in;        if (cin.eof()) break;        in = predo(in);        trans_to_postfix_expression_to_s(in);        cout << calculate_from_postfix_expression() << endl;    }    return 0;}

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