ZOJ 3782 G - Ternary Calculation 水

LINK:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3782

题意：给出3个数和两个符号（+-*/%）

思路：拿到题目还以为是要写波兰表达式，仔细一看固定的数和固定的符号，直接if判断好了

/** @Date    : 2017-03-23-21.31
  * @Author  : Lweleth (SoungEarlf@gmail.com)
  * @Link    : https://github.com/
  * @Version :
  */
#include<bits/stdc++.h>
#define LL long long
#define PII pair
#define MP(x, y) make_pair((x),(y))
#define fi first
#define se second
#define PB(x) push_back((x))
#define MMG(x) memset((x), -1,sizeof(x))
#define MMF(x) memset((x),0,sizeof(x))
#define MMI(x) memset((x), INF, sizeof(x))
using namespace std;
 
const int INF = 0x3f3f3f3f;
const int N = 1e5+20;
const double eps = 1e-8;
 
int main()
{
    int T;
    cin >> T;
    while(T--)
    {
        int a, b, c;
        char o1[2], o2[2];
        int t = 0, ans = 0;
        scanf("%d %s%d %s%d", &a, o1, &b, o2, &c);
        //cout << a << o1 << b << o2 << c << endl;
        if(o1[0] != '*' && o1[0] != '/' && o1[0] != '%' && o2[0] != '+' && o2[0] != '-')
        {
            if(o2[0] == '*')
                t = b * c;
            else if(o2[0] == '/')
                t = b / c;
            else if(o2[0] == '%')
                t = b % c;
            if(o1[0] == '+')
                t = a + t;
            else if(o1[0] == '-')
                t = a - t;
        }
        else
        {
            if(o1[0] == '+')
                t = a + b;
            else if(o1[0] == '-')
                t = a - b;
            else if(o1[0] == '*')
                t = a * b;
            else if(o1[0] == '/')
                t = a / b;
            else if(o1[0] == '%')
                t = a % b;
 
            if(o2[0] == '+')
                t = t + c;
            else if(o2[0] == '-')
                t = t - c;
            else if(o2[0] == '*')
                t = t * c;
            else if(o2[0] == '/')
                t = t / c;
            else if(o2[0] == '%')
                t = t % c;
        }
        //cout << 5 % 8 % 2 << endl;
        printf("%d\n", t);
    }
    return 0;
}