首页 > 代码库 > 【微软编程一小时】题目1 : Arithmetic Expression

【微软编程一小时】题目1 : Arithmetic Expression

2024-07-02 07:59:07   215人阅读

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
2
每次都很傻逼地用 == 去判断String相等，每次都吃亏都不长记性！叫你不长记性！
import java.util.Scanner;


public class Main {

	static int InversionCount ;
	
	public static void main(String[] args) 
	{
		int T,t;
		Scanner jin = new Scanner(System.in);
		T = jin.nextInt();
		jin.nextLine();
		
		int ret = T+1;
		double max_abs = Double.MAX_VALUE;
		
		for (t = 1; t <= T; t++) {

	
			String line = jin.nextLine();
			String[] argStrings = line.split(" ");
			
			//System.out.println(argStrings.length);
			
			double a = Double.parseDouble(argStrings[0]);
			double b = Double.parseDouble(argStrings[2]);
							
			double op_ret ;
			if (argStrings[1].equals("+")) {
				op_ret = a + b;
			}
			else if (argStrings[1].equals("-")) {
				op_ret = a - b;
			}
			else if (argStrings[1].equals("*")) {
				op_ret = a * b;
			}
			else op_ret = a / b;
			
			
			if (Math.abs(op_ret - 9) < max_abs) {
				max_abs = Math.abs(op_ret - 9);
				ret = t;
			}
		}
		System.out.println(ret);
	}
}


【微软编程一小时】题目1 : Arithmetic Expression


联系
我们