【水搜索】#31 A. Worms Evolution_31 a worms evolution-优快云博客

本文介绍了一种算法，用于解决一个关于蠕虫形态的研究问题。教授Vasechkin提出了一个理论，即所有蠕虫通过分裂进化，并需要找到三种特定的蠕虫形态，使得其中一种的长度等于另外两种长度之和。文章提供了一个简单有效的解决方案，采用三层循环来验证这一条件。

Professor Vasechkin is studying evolution of worms. Recently he put forward hypotheses that all worms evolve by division. There are nforms of worms. Worms of these forms have lengths a₁, a₂, ..., a_n. To prove his theory, professor needs to find 3 different forms that the length of the first form is equal to sum of lengths of the other two forms. Help him to do this.

Input

The first line contains integer n (3 ≤ n ≤ 100) — amount of worm's forms. The second line contains n space-separated integers a_i (1 ≤ a_i ≤ 1000) — lengths of worms of each form.

Output

Output 3 distinct integers i j k (1 ≤ i, j, k ≤ n) — such indexes of worm's forms that a_i = a_j + a_k. If there is no such triple, output -1. If there are several solutions, output any of them. It possible that a_j = a_k.

Sample test(s)

input
5
1 2 3 5 7

output
3 2 1

input
5
1 8 1 5 1

output
-1

还亏我想了半天……这道题暴力O (n^3)都能无压力A……

那就直接水掉吧~ FORFORFOR 检测flag~ over~

Code：8k 30ms AC

#include <cmath>   
#include <cstdio>  
#include <string>  
#include <cstring>  
#include <iostream>  
#include <algorithm>
#define FOR(a,b) for(a=b;a<num;a++)
using namespace std;
int list[101];
int main()
{
	int h,i,j,k,num=0;
	scanf("%d",&num);
	memset(list,0,sizeof list);
	FOR(h,0)	scanf("%d",&list[h]);
	int flag=0;
	FOR(i,0)
	{
		FOR(j,i+1)
		{
			FOR(k,j+1)
			{
				flag= 	(list[i]==list[j]+list[k])?1:
						(list[j]==list[k]+list[i])?2:
						(list[k]==list[i]+list[j])?3:0;	
				if(flag)break;
			}
			if(flag)break;
		}
		if(flag)break;
	}
	if(flag)
	{
		int a = (flag==1)?i:(flag==2)?j:k;
		int b = (flag==1)?j:(flag==2)?k:i;
		int c = (flag==1)?k:(flag==2)?i:j;
		printf("%d %d %d",a+1,b+1,c+1);
	}
	else printf("-1");
	return 0;
}