A - Presents
Problem description
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
解题思路:输入n(n表示编号从1到n的学生总数)个数,第i个数表示编号为i的学生送礼物给编号为j的学生,要求输出编号为j的学生收到那个送他礼物的学生编号i,水过!
AC代码:
#include<bits/stdc++.h>
using namespace std;
int main(){
int n,x,t[];
cin>>n;
for(int i=;i<=n;++i){cin>>x;t[x]=i;}
for(int i=;i<=n;++i)
cout<<t[i]<<(i==n?"\n":" ");
return ;
}
A - Presents的更多相关文章
- CodeForces 483B Friends and Presents
Friends and Presents Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I ...
- B. Friends and Presents(Codeforces Round #275(div2)
B. Friends and Presents time limit per test 1 second memory limit per test 256 megabytes input stand ...
- Codeforces483B. Friends and Presents(二分+容斥原理)
题目链接:传送门 题目: B. Friends and Presents time limit per test second memory limit per test megabytes inpu ...
- Codeforces 483B - Friends and Presents(二分+容斥)
483B - Friends and Presents 思路:这个博客写的不错:http://www.cnblogs.com/windysai/p/4058235.html 代码: #include& ...
- CF 483B. Friends and Presents 数学 (二分) 难度:1
B. Friends and Presents time limit per test 1 second memory limit per test 256 megabytes input stand ...
- Codeforces Round #348 (VK Cup 2016 Round 2, Div. 2 Edition) A. Little Artem and Presents 水题
A. Little Artem and Presents 题目连接: http://www.codeforces.com/contest/669/problem/A Description Littl ...
- codefoeces B. Friends and Presents
B. Friends and Presents time limit per test 1 second memory limit per test 256 megabytes input stand ...
- codeforces 669A A. Little Artem and Presents(水题)
题目链接: A. Little Artem and Presents time limit per test 2 seconds memory limit per test 256 megabytes ...
- Codeforces Round #275 (Div. 2) B. Friends and Presents 二分+数学
8493833 2014-10-31 08:41:26 njczy2010 B - Friends and Presents G ...
- A - Alice and the List of Presents (排列组合+快速幂取模)
https://codeforces.com/contest/1236/problem/B Alice got many presents these days. So she decided to ...
随机推荐
- ROS:ubuntu-Ros使用OrbSLAM
一般无误的官方连接:https://github.com/raulmur/ORB_SLAM ubuntu16.04没有多少改变,还是使用kinetic老代替indigo Related Publica ...
- SLAM: 单目视觉SLAM的方案分类《机器人手册》
摘抄知乎上一段有趣的话: 如果你出门问别人『学习SLAM需要哪些基础?』之类的问题,一定会有很热心的大哥大姐过来摸摸你的头,肩或者腰(不重要),一脸神秘地从怀里拿出一本比馒头还厚的<Mu ...
- Step by Step 开发dynamics CRM
这里是作为开发贴的总结. 现在plugin和workflow系列已经终结. 希望这些教程能给想入坑的小伙伴一些帮忙. CRM中文教材不多, 我会不断努力为大家提供更优质的教程. Plugin 开发系列 ...
- BZOJ 3450: Tyvj1952 Easy 数学期望
Code: #include <bits/stdc++.h> #define setIO(s) freopen(s".in","r",stdin) ...
- Python对JSON的操作 day3
下面将为大家介绍如何使用python语言来编码和解码json对象: json串就是一个字符串,json串必须用双引号,不能使用单引号 使用json函数需要导入json库,import json 1.j ...
- 05-Linux系统编程-第02天(文件系统、目录操作、dup2)
1 课程回顾 02-文件存储 文件名不在inode里 而是保存在一个叫dentry的结构体里了 格式化就是指定一组规则 指定对文件的存储及读取的一般方法 linux下主要使用 ext2 ext3 ex ...
- Centos 修改主机名称
Centos 配置主机名称: 1.首先查询一下当前的主机名称 [root@localhost~]# hostnamectl status Static hostname: ****** //永久主机名 ...
- 9.简单理解ES分布式
主要知识点: 1.Elasticsearch对复杂分布式机制的透明隐藏特性 2.Elasticsearch的垂直扩容与水平扩容 3.增减或减少节点时的数据rebalance 4.master节 ...
- 搜索引擎seo优化
<a href="" title="SEO优化"></a> <img src="" alt="SEO ...
- Spring Boot学习总结(3)——SpringBoot魅力所在
使用Java做Web应用开发已经有近20年的历史了,从最初的Servlet1.0一步步演化到现在如此多的框架,库以及整个生态系统.经过这么长时间的发展,Java作为一个成熟的语言,也演化出了非常成熟的 ...