G.Ternary Calculation

Time Limit: 2 Seconds Memory Limit: 65536 KB

Complete the ternary calculation.

Input

There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case:

There is a string in the form of "number₁ operator_anumber₂ operator_b number₃". Each operator will be one of {'+', '-' , '*', '/', '%'}, and each number will be an integer in [1, 1000].

Output

For each test case, output the answer.

Sample Input

5
1 + 2 * 3
1 - 8 / 3
1 + 2 - 3
7 * 8 / 5
5 - 8 % 3

Sample Output

7
-1
0
11
3

Note

The calculation "A % B" means taking the remainder of A divided by B, and "A / B" means taking the quotient. 

/********************************************************************************/

此题也是这届省赛的一道水题，就是普通的四则运算（增加一个取模运算），重点在于优先级，如果前一个运算符是‘+’或者‘-’，后一个运算符是‘*’或‘/‘或’%‘，就要优先计算后两项的结果

#include<stdio.h>
int fun(int m,char z,int n)
{
    int ans=0;
    //printf("%c\n",z);
    switch(z)
    {
        case'+':ans=m+n;break;
        case'-':ans=m-n;break;
        case'*':ans=m*n;break;
        case'/':ans=m/n;break;
        case'%':ans=m%n;break;
    }
    //printf("%d\n",ans);
    return ans;
}
int main()
{
    int t,a,b,c,sum;
    char x,y;
    scanf("%d",&t);
    while(t--)
    {
        sum=0;
        scanf("%d %c %d %c %d",&a,&x,&b,&y,&c);
        //printf("%d %c %d %c %d\n",a,x,b,y,c);
        if(x=='*'||x=='/'||x=='%')
        {
            sum=fun(a,x,b);
            sum=fun(sum,y,c);
        }
        else if(y=='*'||y=='/'||y=='%')
        {
            sum=fun(b,y,c);
            sum=fun(a,x,sum);
        }
        else
        {
            sum=fun(a,x,b);
            sum=fun(sum,y,c);
        }
        printf("%d\n",sum);
    }
    return 0;
}