<cf>Worms Evolution

Worms Evolution

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Professor Vasechkin is studying evolution of worms. Recently he put forward hypotheses that all worms evolve by division. There are n forms 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 nspace-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

想了很久，后来才发觉直接暴力解决就可以轻松过，完全不用任何优化。直接cin然后三重for loop暴力遍历所有组合。我优化了一下，然后发觉跟直接暴力用时完全一样T.T

AC Code:

#include <cstdio>
#include <algorithm>
#include <iostream>
using namespace std;

struct Num
{
    int idx;//index
    int val;//value
}a[1001];

int cmp(const void *xx,const void *yy)
{
    struct Num *x=(Num*)xx;
    struct Num *y=(Num*)yy;
    return x->val-y->val;
}

int main()
{
    int n;
    int i,j,k;
    scanf("%d",&n);
    for(i=1;i<=n;i++)
    {
        scanf("%d",&a[i].val);
        a[i].idx=i;
    }
    qsort(&a[1],n,sizeof(a[1]),cmp);
    for(i=1;i<n;i++)
        for(j=i+1;j<n;j++)
            for(k=j+1;k<=n;k++)
            {
                if(a[k].val==a[i].val+a[j].val)
                {
                    printf("%d %d %d\n",a[k].idx,a[i].idx,a[j].idx);
                    goto loop;//这样退出较快
                }
                else if(a[k].val>a[i].val+a[j].val)
                    break;
            }
    puts("-1");
    loop:
    return 0;
}