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

  1. CodeForces 483B Friends and Presents

     Friends and Presents Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I ...

  2. 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 ...

  3. 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 ...

  4. 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& ...

  5. 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 ...

  6. 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 ...

  7. codefoeces B. Friends and Presents

    B. Friends and Presents time limit per test 1 second memory limit per test 256 megabytes input stand ...

  8. 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 ...

  9. 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 ...

  10. 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

  1. texi格式文件的读取

    使用texi2html可以将texi格式的文件转换成html格式的文件. sudo apt-get install texi2html 在对应目录下 texi2html filename.texi 或 ...

  2. vue.js的ajax和jsonp请求

    首先要声明使用ajax 在 router下边的 Index.js中 import VueResource from 'vue-resource'; Vue.use(VueResource); ajax ...

  3. echarts 中 请求后台改变数据

    function tablenumber() { $.ajax({ type : "get", url : "../res/error.json", dataT ...

  4. TensorFlow实战笔记(17)---TFlearn

    目录: 分布式Estimator 自定义模型 建立自己的机器学习Estimator 调节RunConfig运行时的参数 Experiment和LearnRunner 深度学习Estimator 深度神 ...

  5. 【剑指Offer】26、二叉搜索树与双向链表

      题目描述:   输入一棵二叉搜索树,将该二叉搜索树转换成一个排序的双向链表.要求不能创建任何新的结点,只能调整树中结点指针的指向.   解题思路:   首先要理解此题目的含义,在双向链表中,每个结 ...

  6. CentOS 7.2.1511编译安装Nginx1.10.1+MySQL5.7.15+PHP7.0.11

    准备篇 一.防火墙配置 CentOS 7.2默认使用的是firewall作为防火墙,这里改为iptables防火墙. 1.关闭firewall: systemctl stop firewalld.se ...

  7. poj 3469 最小割模板sap+gap+弧优化

    /*以核心1为源点,以核心2为汇点建图,跑一遍最大流*/ #include<stdio.h> #include<string.h> #include<queue> ...

  8. Timus - 1213 - Cockroaches!

    先上题目: 1213. Cockroaches! Time limit: 1.0 secondMemory limit: 64 MB It's well-known that the most ten ...

  9. codevs——T1219 骑士游历

     http://codevs.cn/problem/1219/  时间限制: 1 s  空间限制: 128000 KB  题目等级 : 黄金 Gold 题解  查看运行结果     题目描述 Desc ...

  10. TCP/IP解析(一):TCP/IP的工作方式

    本文包括下面内容: 1.TCP/IP协议系统 2.OSI模型 3.数据包 4.TCP/IP的交互方式 1.TCP/IP模型的协议层 分为四层: 网络訪问层:提供与物理网络连接的接口.依据硬件的物理地址 ...