简单计算器

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

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;
}