微软编程一小时题目1 : Arithmetic Expression

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

package ArithmeticExpression;

import java.io.IOException;
import java.util.Scanner;

 public class Main {

	public static void main(String[] args) throws IOException {
		long start = System.currentTimeMillis();
		Scanner sc = new Scanner(System.in); 
//		int num  =  Integer.parseInt(sc.nextLine());
		
		//也可以
		int num = sc.nextInt();
		sc.nextLine();
		
		String []strline = new String[num];
		for(int i=0;i<num;i++){
			if(sc.hasNextLine()){
				strline[i]=sc.nextLine();
				System.out.println(strline[i]);
			}
		}
		sc.close();
		
		double []answer = new double[num];
		int count = 0;
		String [] str = new String[3];
		for(int i = 0;i<num;i++){
			str = strline[i].split(" ");
			if(str[1].equals("+"))
				answer[count] = Math.abs(Double.parseDouble(str[0]) + Double.parseDouble(str[2])-9);
			else if(str[1].equals("-"))
				answer[count] = Math.abs(Double.parseDouble(str[0]) - Double.parseDouble(str[2])-9);
			else if(str[1].equals("*"))
				answer[count] = Math.abs(Double.parseDouble(str[0]) * Double.parseDouble(str[2])-9);
			else if(str[1].equals("/"))
				answer[count] = Math.abs(Double.parseDouble(str[0]) / Double.parseDouble(str[2])-9);
			count ++;
		}
		Double min =answer[0];
		int flag = 0;
		for(int j=1;j<num;j++){
			if(min>answer[j]){
				min = answer[j];
				flag = j;
			}
		}
		System.out.println(flag+1);
		long end = System.currentTimeMillis();
		System.out.println(end - start + "ms");
	}

}

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微软编程一小时题目1 : Arithmetic Expression

原文：http://blog.csdn.net/lsp1991/article/details/23556073