2016 UESTC Training for Data Structures N - 秋实大哥搞算数 CDOJ 1074 栈 表达式求值

N -
秋实大哥搞算数
给一个表达式，无括号，保证合法，long long以内，整数运算，求值
栈 表达式求值的经典问题，
首先设置一个开始和结束符号#，他们的优先级是最低的，然后开两个栈，运算符栈和运算数栈，如果当前要放的运算符优先级高的话，就放进去，然后继续模拟，如果当前要放的运算符优先级低的话，就取出栈顶的运算符，和运算数栈顶的两个数，进行运算，然后继续检查，直到栈空或者栈顶运算符优先级低了为止，因为栈最先时放了一个#进去，所以保证合法，只有读入到结束符#时，运算符栈才会被算空
当然， 如果栈顶运算符和要加入的运算符相同，也是优先栈顶运算符计算的
end

代码：

#include <iostream>
#include <vector>
#include <cstring>
#include <cstdio>
#include <string>
#include <algorithm>
#include <vector>
#include <stack>
using namespace std;
#define ll long long
#define maxn 1000005
char str[maxn];
stack<char> oper;
stack<ll> num;
bool prior[5][5] = {
{ 1, 0, 0, 0, 0 },
{ 1, 1, 1, 0, 0 },
{ 1, 1, 1, 0, 0 },
{ 1, 1, 1, 1, 1 },
{ 1, 1, 1, 1, 1 }
};
int idx(char c)
{
int x;
switch (c)
{
case '#':
x = 0; break;
case '+':
x = 1; break;
case '-':
x = 2; break;
case '*':
x = 3; break;
case '/':
x = 4; break;
default:
x = -1; break;
}
return x;
}
ll calcluate(ll a, ll b, char c)
{
ll ans;
switch (c)
{
case '#':
ans = b; break;
case '+':
ans = a + b; break;
case '-':
ans = a - b; break;
case '*':
ans = a*b; break;
case '/':
ans = a / b; break;
}
return ans;
}
int main(){
//freopen("input.txt", "r", stdin);
int T;
scanf("%d", &T);
while (T--)
{
ll nn = 0;
while (!num.empty())
num.pop();
while (!oper.empty())
oper.pop();
memset(str, 0, sizeof(char)*maxn);
scanf("%s", str);
oper.push('#');
int n = strlen(str);
str[n++] = '#';
for (int i = 0; i < n; ++i)
{
if (str[i] >= '0'&&str[i] <= '9')
{
nn = nn * 10 + str[i] - '0';
}
else
{
num.push(nn);
nn = 0;
int x = idx(str[i]);
while (!oper.empty())
{
char c = oper.top();
int y = idx(c);
if (prior[y][x])
{
oper.pop();
ll a = 0, b = 0;
if (!num.empty())
{
a = num.top();
num.pop();
}
if (!num.empty())
{
b = num.top();
num.pop();
}
//printf("%d  %lld %lld %c\n", i, b, a, c);
num.push(calcluate(b, a, c));
}
else
{
oper.push(str[i]);
break;
}
}
}
}
printf("%lld\n", num.top());
}
//while (1);
return 0;
}