ZOJ-3782-Ternary Calculation【11th浙江省赛】

ZOJ-3782-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 operatora number2 operatorb 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.

题目链接：ZOJ-3782

题目大意：简单四则运算

题目思路：模拟

以下是代码：

#include <bits/stdc++.h>
#define mst(a) memset(a,0,sizeof (a))
#define FOR(i,n) for (int i = 0; i < n; i++)
#define INF 1e9
#define eps 1e-10
using namespace std;

typedef long long ll;
int get(char op)
{
    if (op == '+' || op == '-') return 0;
    else return 1;
}
int solve(int a,char op,int b)
{
    if (op == '-') return a - b;
    if (op == '+') return a + b;
    if (op == '*') return a * b;
    if (op == '/') return a / b;
    if (op == '%') return a % b;
}
int main(){
    int t;
    cin >> t;
    while(t--)
    {
        int a,b,c,ans = 0;
        char op1,op2;
        cin >> a >> op1 >> b >> op2 >> c;
        if (get(op1) >= get(op2))
        {
            ans = solve(a,op1,b);
            ans = solve(ans,op2,c);
        }
        else
        {
            ans = solve(b,op2,c);
            ans = solve(a,op1,ans);
        }
        cout << ans << endl;
    } 
    return 0;
}