NYOJ-35-expression evaluate

NYOJ-35-expression evaluate

Http://acm.nyist.net/JudgeOnline/problem.php? Pid = 35

 

A good question: Using stacks to calculate the expression value

Rules for arithmetic operations: 1. First multiplication and division, then addition and subtraction; 2. From left to right; 3. Parentheses, and then parentheses

Note that you can use the atof function to convert a string to a floating point number.

# Include <stdio. h> # include <string. h> # include <stdlib. h> int map [7] [7] = // The priority relationship between operators. The value 100 indicates that no occurrence occurs {,-1,-1, 1}, {,-1,-1,-, 1}, {,-, 1}, {,-, 1 }, {-1,-1,-1,-1,-1,-100,}, {, 1}, {-1,-1,-1, -1,-1,100, 0 }}; int cam (char c) {switch (c) {case '+': return 0; case '-': return 1; case '*': return 2; case '/': return 3; case '(': return 4; case ')': return 5; case '#': return 6 ;}} double sol (double X, char c, double y) {switch (c) {case '+': return x + y; case '-': return x-y; case '*': return x * y; case '/': return x/y;} int z (char c) {if ('0' <= c & c <= '9' | c = '. ') return 1; if (c = '') return-1; return 0;} char str [1005]; char optr [1005]; double opnd [1005]; int main () {int t1, t2, k, len; char ch, zz; int temp1, temp2; double a, B; int t; scanf ("% d ", & t); getchar (); while (t --) {gets (str); len = strlen (str); str [len-1] = '#'; // process equal to t1 = T2 = k = 0; optr [t1 ++] = '#'; ch = str [k ++]; while (ch! = '#' | Optr [t1-1]! = '#') {If (z (ch) = 1) // operand into Stack {opnd [t2 ++] = atof (& str [k-1]); // convert the string to a floating point while (z (str [k]) = 1) k ++; ch = str [k ++];} else if (z (ch) =-1) ch = str [k ++]; else {temp1 = cam (optr [t1-1]); temp2 = cam (ch ); if (map [temp1] [temp2] =-1) // The top element of the stack has a low priority {optr [t1 ++] = ch; ch = str [k ++];} else if (map [temp1] [temp2] = 0) // remove the brackets and accept the next character {t1 --; ch = str [k ++];} else // return the stack and return the calculation result {zz = optr [-- t1]; a = opnd [-- t2]; B = opnd [-- t2]; opnd [t2 ++] = sol (B, zz, a) ;}} printf ("%. 2lf \ n ", opnd [0]);} return 0 ;}