SGU 299 Triangle（大数）

寻找可构成三角形的三边

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本题要求从N个整数中找出三个数，这三数能作为三角形的三边长度。输入包括N个整数，输出则是这三个整数或0表示不存在这样的组合。采用排序和两层循环来解决此问题。

Triangle

Description

It is well known that three segments can make a triangle if and only if the sum of lengths of any two of them is strictly greater than the length of the third one. Professor Vasechkin has N segments. He asked you, if you could find at least one set of three segments among them which can be used by professor to make a triangle.

Input

The first line of the input contains the only integer number N (3≤ N≤ 1000). The following N lines contain the length of segments professor has. The length of any segment is the integer number from 1 to 10 ⁵⁰⁰.

Output

Write to the output the length of segments requested by the professor — three numbers delimited by spaces. Write three zeros if there are no such three segments.

Sample Input

sample input	sample output
7 1 2 6 4 8 100 73	8 4 6

解题思路：

比较坑的位置就是：

public class Solution{  

}

AC代码：

import java.math.BigInteger;
import java.util.Arrays;
import java.util.Scanner;

public class Solution{  
    public static void main(String args[]){  
    	Scanner sca = new Scanner(System.in);
        while(sca.hasNext()){
        	int n = sca.nextInt();
        	BigInteger[] a = new BigInteger[n];
            boolean flag = true;
            for(int i = 0; i < n; i++)
            	a[i] = sca.nextBigInteger();
            Arrays.sort(a);
            for(int i = 0; i < n && flag; i++){
            	for(int j = i+1; j < n-1 && flag; j++){
        			if((a[i].add(a[j])).compareTo(a[j+1]) > 0){
        				flag = false;
        				System.out.println(a[i]+" "+a[j]+" "+a[j+1]);
        			}
            	}
            }
            if(flag)
            	System.out.println("0 0 0");
        }
    }  
}