Problem Description
读入一个只包含 +, -, *, / 的非负整数计算表达式,计算该表达式的值。
Input
测试输入包含若干测试用例,每个测试用例占一行,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。当一行中只有0时输入结束,相应的结果不要输出。
Output
对每个测试用例输出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
string 读入
#include<stdio.h>
#include<iostream>
#include<cstring>
#include<vector>
#include<algorithm>
#include<string>
#include<sstream>
#include<stack>
using namespace std;
string s;
stack<double> a;
stack<char>b;
double x, y;
char c, z;
bool check(char x, char y)
{
if (x == '*' || x == '/') return true;
if (y == '+' || y == '-') return true;
return false;
}
int main()
{
while (getline(cin, s), s != "0")
{
s = s + ' ' + 'A';
stringstream ss(s);
while (ss >> x >> c)
{
a.push(x);
while (!b.empty() && (c == 'A' || check(b.top(), c)))
{
x = a.top(); a.pop();
y = a.top(); a.pop();
z = b.top(); b.pop();
switch (z){
case '+':a.push(x + y); break;
case '-':a.push(y - x); break;
case '*':a.push(x * y); break;
case '/':a.push(y / x); break;
}
}
if (c != 'A') b.push(c);
}
printf("%.2f\n", a.top());
while (!a.empty()) a.pop();
}
return 0;
}
char 读入
#include<stdio.h>
#include<string.h>
#include<stack>
using namespace std;
int i;
double x, y;
char s[250], z;
bool check(char x, char y)
{
if (x == '*' || x == '/') return true;
if (y == '+' || y == '-') return true;
return false;
}
int main()
{
while (gets(s), strcmp(s, "0") != 0)
{
stack<char> b;
stack<double> a;
for (i = 0; s[i]; i++)
{
if (s[i] == ' ') continue;
if (s[i] >= '0'&&s[i] <= '9')
{
x = 0;
while (s[i] >= '0'&&s[i] <= '9')
{
x = x * 10 + s[i] - '0';
i++;
}
i--;
a.push(x);
}
if (s[i] < '0' || s[i] > '9' || !s[i + 1])
{
while (!b.empty() && (!s[i + 1] || check(b.top(), s[i])))
{
x = a.top(); a.pop();
y = a.top(); a.pop();
z = b.top(); b.pop();
switch (z){
case '+':a.push(x + y); break;
case '-':a.push(y - x); break;
case '*':a.push(x * y); break;
case '/':a.push(y / x); break;
}
}
b.push(s[i]);
}
}
printf("%.2lf\n", a.top());
}
return 0;
}
注意头文件要用stdio.h,用cstdio一直wa,搞不懂。