PTA波兰表达式

一、题目描述

　　

 

 二、解题思路

　　弄个string的栈，把这些全部存进栈，然后从栈顶一直往下走，碰到两个数字和一个操作符就进行一次操作，再入栈就行。（注意除号和减号不要写反了，不然跟我一样调一个小时）

三、代码实现

 1 #include "bits/stdc++.h"
 2 #define PII pair<int,int>
 3 #define rep(i,z,n) for(int i = z;i <= n; i++)
 4 #define per(i,n,z) for(int i = n;i >= z; i--)
 5 #define ll long long
 6 #define db double
 7 #define vi vector<int>
 8 #define debug(x) cerr << "!!!" << x << endl;
 9 using namespace std;
10 inline ll read()
11 {
12     ll s,r;
13     r = 1;
14     s = 0;
15     char ch = getchar();
16     while(ch < '0' || ch > '9'){
17         if(ch == '-')
18             r = -1;
19         ch = getchar();
20     }
21     while(ch >= '0' && ch <= '9'){
22         s = (s << 1) + (s << 3) + (ch ^ 48);
23         ch = getchar();
24     }
25     return s * r;
26 }
27 inline void write(ll x)
28 {
29     if(x < 0) putchar('-'),x = -x;
30     if(x > 9) write(x / 10);
31     putchar(x % 10 + '0');
32 }
33 stack <string> que;
34 stack <double> num;
35 stack <string> c;
36 int main()
37 {
38     string s;
39     string temp = "";
40     getline(cin,s);
41     // reverse(s.begin(),s.end());
42     // cout << s << endl
43     for(auto &u:s){
44         if(u == ' '){
45             que.push(temp);
46             temp = "";
47             continue;
48         }
49         temp += u;
50     }
51     if(temp.size() != 0)
52         que.push(temp);
53     string code;
54     while(!que.empty()){
55         string temp = que.top();
56         que.pop();
57         if(temp.size() == 1)
58             c.push(temp);
59         if(temp.size() > 1){
60             double k = stod(temp);
61             num.push(k);    
62         }
63         if(num.size() >= 2 && !c.empty()){
64             double k;
65             code = c.top();
66             c.pop();
67             double val1 = num.top();
68             num.pop();
69             double val2 = num.top();
70             num.pop();
71             if(code == "+")
72                 k = val1 + val2;
73             else if(code == "-")
74                 k = val1 - val2;
75             else if(code == "*")
76                 k = val1 * val2;
77             else 
78                 k = val1 / val2;
79             // cout << k << endl;
80             num.push(k);
81         }
82     }
83     cout << fixed << setprecision(6) << num.top() << endl;
84     return 0;
85 }

 

posted @ 2022-03-10 21:56  scannerkk  阅读(99)  评论(0)    收藏  举报