HDU1237 简单计算器 栈+逆波兰式

Posted 2020-09-19 jzssuanfa

简单计算器

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 11955    Accepted Submission(s): 3896

Problem Description

读入一个仅仅包括 +, -, *, / 的非负整数计算表达式，计算该表达式的值。

 

Input

測试输入包括若干測试用例。每一个測试用例占一行，每行不超过200个字符，整数和运算符之间用一个空格分隔。没有非法表达式。当一行中仅仅有0时输入结束。对应的结果不要输出。

 

Output

对每一个測试用例输出1行，即该表达式的值，精确到小数点后2位。

 

Sample Input

1 + 2
4 + 2 * 5 - 7 / 11
0

 

Sample Output

3.00
13.36

关键地方是在把中缀式转换成后缀式时要保持符号栈从顶開始严格递减，否则先出栈，再进栈。

#include <stdio.h>
#include <string.h>
char str[202], buf[202], sign[202]; //buf存储逆波兰式
double stack[202], a;
int len, n, id, id2, id3, id4;

double perform(double x, double y, char ch){
	if(ch == ‘*‘) return x * y;
	if(ch == ‘/‘) return x / y;
	if(ch == ‘+‘) return x + y;
	return x - y;
}

void check(char ch){
	buf[id2++] = ‘ ‘;
	if(ch == ‘*‘ || ch == ‘/‘){	
		while(id3 && (sign[id3-1] == ‘*‘ || sign[id3-1] == ‘/‘)) 
			buf[id2++] = sign[--id3];
		sign[id3++] = ch;
		return;
	}
	while(id3) buf[id2++] = sign[--id3];
	sign[id3++] = ch;
}

int main(){	
	while(gets(str)){
		len = strlen(str);
		if(len == 1 && str[0] == ‘0‘) break;
		id = id2 = id3 = id4 = 0;
		for(int i = 0; i < len; ++i){
			if(str[i] == ‘ ‘) continue;
			if(str[i] >= ‘0‘ && str[i] <= ‘9‘){
				buf[id2++] = str[i];
			}else check(str[i]);
		}
		while(id3) buf[id2++] = sign[--id3];
		//for(int i = 0; i < id2; ++i) putchar(buf[i]);
	 	for(int i = 0; i < id2; ++i){
			if(buf[i] == ‘ ‘) continue;
			if(buf[i] >= ‘0‘ && buf[i] <= ‘9‘){
				sscanf(buf + i, "%lf%n", &stack[id4++], &n);
				i += n - 1;
			}else stack[id4-2] = perform(stack[id4-2], stack[id4-1], buf[i]), --id4;			
		}
		
		printf("%.2lf\n", stack[0]);
	}
	return 0;
}

2014-11-3 21:55:30更新

#include <stdio.h>
#include <string.h>

#define maxn 1000

char buf[maxn], out[maxn], stack[maxn];
int id, id1, ida;
double A[maxn];

int level(char ch) {
	if(ch == ‘+‘ || ch == ‘-‘) return 1;
	return 2;
}

void check(char ch) {
	while(id1 && level(stack[id1-1]) >= level(ch)) {
		out[id++] = stack[--id1];
	}
	stack[id1++] = ch;
}

double cal(double a, double b, char ch) {
	if(ch == ‘-‘) return a - b;
	if(ch == ‘+‘) return a + b;
	if(ch == ‘*‘) return a * b;
	return a / b;
}

int main() {
	int i, n;
	bool sign;
	double a;
	while(gets(buf)) {
		if(strlen(buf) == 1 && buf[0] == ‘0‘)
			break;
		id = id1 = 0; sign = 0;
		for(i = 0; buf[i]; ++i) {
			if(buf[i] == ‘ ‘) continue;
			if(buf[i] >= ‘0‘ && buf[i] <= ‘9‘ || buf[i] == ‘.‘) {
				if(sign) {
					out[id++] = ‘ ‘;
					sign = 0;
				}
				out[id++] = buf[i];
			}
			else sign = 1, check(buf[i]);
		}

		while(id1) {
			out[id++] = stack[--id1];
		}

		for(i = ida = 0; i < id; ++i) {
			if(out[i] == ‘ ‘) continue;
			if(out[i] >= ‘0‘ && out[i] <= ‘9‘ || out[i] == ‘.‘) {
				sscanf(out + i, "%lf%n", &a, &n);
				A[ida++] = a; i += n - 1;
			} else A[ida-2] = cal(A[ida-2], A[ida-1], out[i]), --ida;
		}

		printf("%.2lf\n", A[0]);
	}
	return 0;
}