后缀表达式

题目1 只求后缀表达式的值

后缀表达式的值 （改进了链接里的写法）

输入

16 9 4 3 +*-@

输出

-47

代码

#include<iostream> #include<cstdio> #include<cstdlib> #include<string> #include<cstring> #include<cmath> #include<ctime> #include<algorithm> #include<utility> #include<stack> #include<queue> #include<vector> #include<set> #include<map> #include<stdlib.h> #define PI acos(-1.0) #define E 1e-9 #define INF 0x3f3f3f3f #define LL long long #pragma warning(disable:4996) const int MOD = 1000000007; const int N = 10000 + 5; const int dx[] = { -1,1,0,0 }; const int dy[] = { 0,0,-1,1 }; using namespace std; string str; stack<float>st; int cal(int a, int b, char c) { if (c == '+')return a + b; if (c == '-')return a - b; if (c == '*')return a * b; if (c == '/')return a / b; return -1; } int main() { char multNum[10]; strcpy(multNum, "\0"); char cur[2]; float data; getline(cin, str); for (int i = 0; i < str.length(); i++) { if (str[i] == '+' || str[i] == '-'|| str[i] == '*' || str[i] == '/') { LL a = st.top(); st.pop(); LL b = st.top(); st.pop(); st.push(cal(b, a, str[i])); } else if(str[i]=='@') { i = str.length(); } else { while (str[i] != ' ') { cur[0] = str[i]; cur[1] = '\0'; strcat(multNum, cur); i++; } data = (float)atof(multNum); cout << data << endl; strcpy(multNum, "\0"); st.push(data); } } cout << st.top() << endl; }

题目2 中缀表达式转为后缀表达式，再求值

输入

2 1+2*3-4/5# (66+(((11+22)*2-33)/3+6)*2)-45.6789#

输出

6.2000 54.3211

代码

#include<iostream> #include<stack> #include<string> #include<iomanip> #pragma warning(disable:4996) using namespace std; typedef int Status; #define OPSETSIZE 7 float cal(float a, float b, char c) { if (c == '+')return a + b; if (c == '-')return a - b; if (c == '*')return a * b; if (c == '/')return a / b; } char Prior[7][7] = { '>','>','<','<','<','>','>', '>','>','<','<','<','>','>', '>','>','>','>','<','>','>', '>','>','>','>','<','>','>', '<','<','<','<','<','=',' ', '>','>','>','>',' ','>','>', '<','<','<','<','<',' ','=', }; char fuhao[OPSETSIZE] = { '+','-','*','/','(',')','#' }; char diff(char a,char b) { int x, y; for (int i = 0; i < OPSETSIZE; i++) { if (a == fuhao[i])x = i; if (b == fuhao[i])y = i; } return Prior[x][y]; } Status IN(char c) { for (int i = 0; i < OPSETSIZE; i++) { if (c == fuhao[i])return 1; } return 0; } float culExp(string exp) { stack<char>fuhao; stack<float>num; char resNum[20]; strcpy(resNum, "\0"); char curNum[2]; float data; fuhao.push('#'); int i = 0; while(exp[i]!='#'||fuhao.top()!='#') { if (IN(exp[i])) { switch (diff(fuhao.top(), exp[i])) { case '<': fuhao.push(exp[i++]); break; case '=': fuhao.pop(); i++; break; case '>': float a = num.top(); num.pop(); float b = num.top(); num.pop(); num.push(cal(b, a, fuhao.top())); fuhao.pop(); break; } } else { curNum[0] = exp[i]; curNum[1] = '\0'; strcat(resNum, curNum); i++; if (IN(exp[i])) { data = (float)atof(resNum); strcpy(resNum, "\0"); num.push(data); } } } return num.top(); } int main() { int t; cin >> t; string exp; while (t--) { cin >> exp; cout << fixed << setprecision(4)<<culExp(exp) << endl; } }