微软编程一小时比赛--题目1 : Arithmetic Expression

我不是女神ヾ 2022-08-26 12:13 155阅读 0赞

### 题目1 : Arithmetic Expression ###

时间限制: 2000ms

单点时限: 200ms

内存限制: 256MB

### 描述 ###

Given N arithmetic expressions, can you tell whose result is closest to 9?

### 输入 ###

Line 1: N (1 <= N <= 50000).  
Line 2..N+1: Each line contains an expression in the format of "a op b" where a, b are integers (-10000 <= a, b <= 10000) and op is one of addition (+), subtraction (-), multiplication (\*) and division (/). There is no "divided by zero" expression.

### 输出 ###

The index of expression whose result is closest to 9. If there are more than one such expressions, output the smallest index.

样例输入

4
    901 / 100
    3 * 3
    2 + 6
    8 - -1

样例输出

解析：比赛时间太短了，这题明显是水题，但是好像当时自己不在状态，一直调试不出来，还是感觉自己太水了，弄的现在调试出来了没法提交了也不知道对还是不对，这道题没涉及什么算法，主要是在求除法的时候有些小问题，其他应该就没什么了哈！

#include <iostream>
    using std::endl;
    using std::cin;
    using std::cout;
    //operation的函数
    double operation(int a , char op , int b)
    {
    	double result;
    	if(op == '+')
    	{
    		result = a + b;
    	}else if(op == '-')
    	{
    		result = a - b;
    	}else if(op == '*')
    	{
    		result = a * b;
    	}else{
    		result = (double)a / (double)b;
    	}
    	return result;
    }
    int main(void)
    {
    	int N;
    	cin >> N;
    	//distance 记录每组表达式与目标值的接近程度
    	double distance = 1000000000;
    	//保存最后的接近的索引
    	int index;
    	int cnt=1;
    	while(cnt<=N)
    	{
    		int a,b;
    		char op;
    		//输入每组表达式
    		cin >> a >> op >> b;
    		double res;
    		res = operation(a , op , b);
    		//计算与目标值的接近程度
    		double temp = res - 9;
    		//如果小于0则取反
    		if(temp < 0)
    		{
    			temp = -temp;
    		}
    		//比较当前是否最接近如果当前最接近的话则进行替换
    		if(temp < distance)
    		{
    			distance = temp;
    			index = cnt;
    		}
    		cnt++;
    	}
    	cout << index << endl;
    	return 0;
    }