Codeforces Beta Round #97 (Div. 2)

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最新推荐文章于 2024-09-25 11:53:33 发布

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分类专栏： Codeforces 文章标签： output input each up c

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本文详细描述了小彼得如何利用新年派对的机会，通过观察朋友们之间的礼物交换过程，来理解并记录每一份礼物的来源。文章探讨了在不参与交换的情况下，如何通过简单的调查和统计，找出每份礼物的发送者。

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Little Petya very much likes gifts. Recently he has received a new laptop as a New Year gift from his mother. He immediately decided to give it to somebody else as what can be more pleasant than giving somebody gifts. And on this occasion he organized a New Year party at his place and invited n his friends there.

If there's one thing Petya likes more that receiving gifts, that's watching others giving gifts to somebody else. Thus, he safely hid the laptop until the next New Year and made up his mind to watch his friends exchanging gifts while he does not participate in the process. He numbered all his friends with integers from 1 to n. Petya remembered that a friend number i gave a gift to a friend number p_i. He also remembered that each of his friends received exactly one gift.

Now Petya wants to know for each friend i the number of a friend who has given him a gift.

<p< p=""><p< p="">

Input

<p< p="">

The first line contains one integer n (1 ≤ n ≤ 100) — the quantity of friends Petya invited to the party. The second line contains n space-separated integers: the i-th number is p_i — the number of a friend who gave a gift to friend number i. It is guaranteed that each friend received exactly one gift. It is possible that some friends do not share Petya's ideas of giving gifts to somebody else. Those friends gave the gifts to themselves.

<p< p=""><p< p="">

Output

<p< p="">

Print n space-separated integers: the i-th number should equal the number of the friend who gave a gift to friend number i.

<p< p=""><p< p="">

Sample test(s)

<p< p=""><p< p="">

input

4
2 3 4 1

output

4 1 2 3

input

3
1 3 2

output

1 3 2

input

2
1 2

output

1 2

#include<cstdio>

int main()
{
    int a, b, c, d;
    int n, from, hash[200];
    scanf("%d",&n);
    for ( int i = 1; i <= n; i++ )
    {
        scanf("%d",&from);
        hash[from] = i;
    }
    for ( int i = 1; i <= n; i++ )
        printf("%d ",hash[i]);
    printf("\n");
    return 0;
}