Presents

简介: Presents

文章目录

一、 Presents

总结


一、 Presents

本题链接:Presents


题目:

A. Presents

time limit per test2 seconds

memory limit per test256 megabytes

inputstandard input

outputstandard output

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 pi. 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.


Input

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 pi — 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.


Output

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


Examples

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


本博客给出本题截图:

image.png

image.png

题意:输入n个数字,第i个数字代表第i个人把自己的礼物送给了谁,要求按位输出每一个人收到的是哪一个人送出的礼物

AC代码

#include <iostream>
using namespace std;
const int N = 110;
int p[N], pre[N];
int main()
{
    int n;
    cin >> n;
    for (int i = 1; i <= n; i ++ )
    {
        cin >> p[i];
        pre[p[i]] = i;
    }
    for (int i = 1; i <= n; i ++ )
        cout << pre[i] << ' ';
    return 0;
}

总结

水题,不解释


目录
相关文章
|
4天前
|
云安全 人工智能 算法
以“AI对抗AI”,阿里云验证码进入2.0时代
三层立体防护,用大模型打赢人机攻防战
1315 4
|
4天前
|
机器学习/深度学习 安全 API
MAI-UI 开源:通用 GUI 智能体基座登顶 SOTA!
MAI-UI是通义实验室推出的全尺寸GUI智能体基座模型,原生集成用户交互、MCP工具调用与端云协同能力。支持跨App操作、模糊语义理解与主动提问澄清,通过大规模在线强化学习实现复杂任务自动化,在出行、办公等高频场景中表现卓越,已登顶ScreenSpot-Pro、MobileWorld等多项SOTA评测。
660 3
|
5天前
|
人工智能 Rust 运维
这个神器让你白嫖ClaudeOpus 4.5,Gemini 3!还能接Claude Code等任意平台
加我进AI讨论学习群,公众号右下角“联系方式”文末有老金的 开源知识库地址·全免费
|
11天前
|
编解码 人工智能 自然语言处理
⚽阿里云百炼通义万相 2.6 视频生成玩法手册
通义万相Wan 2.6是全球首个支持角色扮演的AI视频生成模型,可基于参考视频形象与音色生成多角色合拍、多镜头叙事的15秒长视频,实现声画同步、智能分镜,适用于影视创作、营销展示等场景。
766 6
|
8天前
|
物联网 API UED
Qwen-Image-Edit-2511来啦!角色一致性再提升,LoRA能力内置
Qwen-Image-Edit-2511发布!提升角色与多人合照一致性,集成Lora打光、新视角生成,增强工业设计与几何推理能力。已开源,支持魔搭、QwenChat免费体验,本地部署可获最佳效果。
465 3