Just determine whether an algebraic expression can always simplify to zero.

Input

The first line contains a single integer T, indicating the number of test cases.
Each test case starts with an integer N, the number of tokens that describes a formula. The next N tokens describe a formula in reverse polish notation.
The notation works as follows. There is a stack that begins empty, and only the following commands manipulate the contents of the stack:
1. “x” pushes the variable x to the stack.
2. “sin”, “cos”, and “tan” replace the top element of the stack with its sin, cos, and tan, respectively.
3. “+”, “-”, and “*” replace the top two elements of the stack (a on top, followed by b) with their sum(b + a), difference (b − a), and product (b ∗ a), respectively.
You may assume that the input is valid, and results in a single item on the stack, which is the desired expression. The length of a line will be at most 300 characters. Note function arguments can contain functions.
 

Output

For each test case, output the case number first, then “Yes” if the expression is always zero, otherwise output “No”.
 

Sample Input

2
3 x sin sin
15 x sin x sin * x cos x cos * + x * x -
 

Sample Output

Case 1: No
Case 2: Yes
题意：判断表达式是否恒为0
思路：因为三角函数都是周期性变化的，所以我们可以枚举x来计算，但是精度自己内心也不确定，就按0.0001的精度累加，也A了
  
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <stack>
#include <math.h>
using namespace std;

#define pi acos(-1.0)

stack<double> S;
char str[305][10];

int main()
{
    int t,n,i,j,cas = 1;
    int flag;
    double x,y,z;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(i = 0; i<n; i++)
            scanf("%s",str[i]);
        flag = 1;
       for(x = -2.0; x<=2.0; x+=0.0001)//枚举x
        {
            for(j = 0; j<n; j++)
            {
                if(!strcmp(str[j],"x"))
                    S.push(x);
                else if(!strcmp(str[j],"sin"))
                {
                    y = S.top();
                    S.pop();
                    y = sin(y);
                    S.push(y);
                }
                else if(!strcmp(str[j],"cos"))
                {
                    y = S.top();
                    S.pop();
                    y = cos(y);
                    S.push(y);
                }
                else if(!strcmp(str[j],"tan"))
                {
                    y = S.top();
                    S.pop();
                    y = tan(y);
                    S.push(y);
                }
                else if(!strcmp(str[j],"+"))
                {
                    y = S.top();
                    S.pop();
                    z = S.top();
                    S.pop();
                    y = y+z;
                    S.push(y);
                }
                else if(!strcmp(str[j],"-"))
                {
                    y = S.top();
                    S.pop();
                    z = S.top();
                    S.pop();
                    y = z-y;
                    S.push(y);
                }
                else if(!strcmp(str[j],"*"))
                {
                    y = S.top();
                    S.pop();
                    z = S.top();
                    S.pop();
                    y = y*z;
                    S.push(y);
                }
            }
            y = S.top();
            S.pop();
            if(fabs(y)<1e-8 && S.empty())
                continue;
            else
            {
                flag = 0;
                break;
            }
        }
        printf("Case %d: ",cas++);
        if(flag)
            printf("Yes\n");
        else
            printf("No\n");
    }

    return 0;
}