Description
基于任意给定的表达式串（包含的运算有加(+)、减(-)、乘(*)、除(/)、圆括号等，例如：输入3.4+5.6*(4.2-1），求其后缀表达式，并求表达式的计算结果。
Input
有多组输入数据，每一行一组输入，请处理到文件结束（EOF）。
每行一个表达式串（最多1000个字符），中间没有空格等其他无关字符。
Output
输出为三行。第一行输出每组的 case，第二行是所求的后缀表达式，第三行是所求的表达式的值。只有每个数字后面会有空格，没有多余空格输出，后缀表达式中的数字按给出字符串中的数字 原样输出，表达式的值中不要有无意义的0出现，每组输出后边都有一个空行。格式见Sample。
Sample Input
5+10*(5+6)-10
5.0+10.00*(5+6.000)-10.00000
Sample Output
Case #1: 5 10 5 6 +*+10 -
The answer is 105.
Case #2: 5.0 10.00 5 6.000 +*+10.00000 -
The answer is 105.
HINT
考察知识点：栈, 时间复杂度O(n)，空间复杂度O(n)
Append Code
析：用栈来计算，按照转化规则，要么自己推一下也是可以，不会看这http://www.cnblogs.com/dwtfukgv/articles/5875526.html
其他的就很简单了。
代码如下：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e3 + 5;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
return r >= 0 && r < n && c >= 0 && c < m;
}
stack<string> num;
stack<char> mark;
char s[maxn];
vector<string> ans;
int priv[maxn];
double calc(double a, double b, char op){
if(op == '+') return a + b;
if(op == '-') return a - b;
if(op == '*') return a * b;
if(op == '/') return a / b;
}
double solve(){
string str = (string)s;
stack<double> num;
priv['+'] = priv['-'] = 3;
priv['*'] = priv['/'] = 2;
priv['('] = 10;
double x, y, t = 0;
char last = 0;
for(int i = 0; i < n; ++i){
if(isdigit(s[i])){
num.push(atof(str.c_str()+i));
for(; i+1 < n && isdigit(str[i+1]); i++);
if(i+1 < n && str[i+1] == '.')
for(i++; i+1 < n && isdigit(str[i+1]); i++);
}
else if(str[i] == '(') mark.push(str[i]);
else if(str[i] == ')'){
while(mark.top() != '('){
y = num.top(); num.pop();
x = num.top(); num.pop();
char op = mark.top(); mark.pop();
num.push(calc(x, y, op));
}
mark.pop();
}
else if(str[i] == '-' && (last == 0 || last == '(')){
num.push(0.0);
mark.push('-');
}
else if(priv[str[i]] > 0){
while(mark.size() > 0 && priv[str[i]] >= priv[mark.top()]){
y = num.top(); num.pop();
x = num.top(); num.pop();
char op = mark.top(); mark.pop();
num.push(calc(x, y, op));
}
mark.push(str[i]);
}
else continue;
last = str[i];
}
while(mark.size() > 0){
y = num.top(); num.pop();
x = num.top(); num.pop();
char op = mark.top(); mark.pop();
num.push(calc(x, y, op));
}
return num.top();
}
int main(){
int kase = 0;
while(scanf("%s", s) == 1){
printf("Case #%d:\n", ++kase);
n = strlen(s);
ans.clear();
for(int i = 0; i < n; ++i){
if(isdigit(s[i])){
string ss = "";
int j;
for(j = i; j < n; ++j){
if(isdigit(s[j]) || s[j] == '.') ss += s[j];
else break;
}
i = j-1;
num.push(ss);
}
else if(s[i] == '('){
mark.push(s[i]);
}
else if(s[i] == ')'){
while(true){
char ch = mark.top(); mark.pop();
if(ch == '(') break;
string s1; s1.push_back(ch);
num.push(s1);
}
}
else{
if(mark.empty()){ mark.push(s[i]); continue; }
while(!mark.empty()){
char ch = mark.top();
string s1; s1.push_back(ch);
if(s1[0] == '('){ mark.push(s[i]); break; }
if((s1[0] == '+' || s1[0] == '-') && (s[i] == '*' || s[i] == '/')){ mark.push(s[i]); break; }
else{
num.push(s1);
mark.pop();
}
}
if(mark.empty()){ mark.push(s[i]); continue; }
}
}
while(!mark.empty()){
char ch = mark.top();mark.pop();
string s1; s1.push_back(ch);
num.push(s1);
}
while(!num.empty()) ans.push_back(num.top()), num.pop();
for(int i = ans.size()-1; i >= 0; --i){
if(isdigit(ans[i][0])) cout << ans[i] << " ";
else cout << ans[i];
}
cout << endl;
printf("The answer is %g.\n\n", solve());
}
return 0;
}
