B. The Best Gift
time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Emily's birthday is next week and Jack has decided to buy a present for her. He knows she loves books so he goes to the local bookshop, where there are n books on sale from one of m genres.

In the bookshop, Jack decides to buy two books of different genres.

Based on the genre of books on sale in the shop, find the number of options available to Jack for choosing two books of different genres for Emily. Options are considered different if they differ in at least one book.

The books are given by indices of their genres. The genres are numbered from 1 to m.

Input

The first line contains two positive integers n and m (2 ≤ n ≤ 2·105, 2 ≤ m ≤ 10) — the number of books in the bookstore and the number of genres.

The second line contains a sequence a1, a2, ..., an, where ai (1 ≤ ai ≤ m) equals the genre of the i-th book.

It is guaranteed that for each genre there is at least one book of that genre.

Output

Print the only integer — the number of ways in which Jack can choose books.

It is guaranteed that the answer doesn't exceed the value 2·109.

Sample test(s)
input
4 3
2 1 3 1
output
5
input
7 4
4 2 3 1 2 4 3
output
18
Note

The answer to the first test sample equals 5 as Sasha can choose:

  1. the first and second books,
  2. the first and third books,
  3. the first and fourth books,
  4. the second and third books,
  5. the third and fourth books.

记录每个种类的本数,然后两两相乘就好了

#include<stdio.h>
//#include<bits/stdc++.h>
#include<string.h>
#include<iostream>
#include<math.h>
#include<sstream>
#include<set>
#include<queue>
#include<map>
#include<vector>
#include<algorithm>
#include<limits.h>
#define inf 0x3fffffff
#define INF 0x3f3f3f3f
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define ULL unsigned long long
using namespace std;
int i,j;
int t;
int n,m;
int num;
map<__int64,__int64> q;
map<__int64,__int64>::iterator it;
map<__int64,__int64>::iterator itt;
int main()
{
__int64 sum=0;
cin>>n>>m;
for(i=0;i<n;i++)
{
cin>>num;
q[num]++;
}
for(it=q.begin();it!=q.end();it++)
{
for(itt=q.begin();itt!=q.end();itt++)
{
if(it->first!=itt->first)
{
sum+=(it->second)*(itt->second);
}
}
}
cout<<sum/2<<endl;
return 0;
}

  

Educational Codeforces Round 3 B的更多相关文章

  1. [Educational Codeforces Round 16]E. Generate a String

    [Educational Codeforces Round 16]E. Generate a String 试题描述 zscoder wants to generate an input file f ...

  2. [Educational Codeforces Round 16]D. Two Arithmetic Progressions

    [Educational Codeforces Round 16]D. Two Arithmetic Progressions 试题描述 You are given two arithmetic pr ...

  3. [Educational Codeforces Round 16]C. Magic Odd Square

    [Educational Codeforces Round 16]C. Magic Odd Square 试题描述 Find an n × n matrix with different number ...

  4. [Educational Codeforces Round 16]B. Optimal Point on a Line

    [Educational Codeforces Round 16]B. Optimal Point on a Line 试题描述 You are given n points on a line wi ...

  5. [Educational Codeforces Round 16]A. King Moves

    [Educational Codeforces Round 16]A. King Moves 试题描述 The only king stands on the standard chess board ...

  6. Educational Codeforces Round 6 C. Pearls in a Row

    Educational Codeforces Round 6 C. Pearls in a Row 题意:一个3e5范围的序列:要你分成最多数量的子序列,其中子序列必须是只有两个数相同, 其余的数只能 ...

  7. Educational Codeforces Round 9

    Educational Codeforces Round 9 Longest Subsequence 题目描述:给出一个序列,从中抽出若干个数,使它们的公倍数小于等于\(m\),问最多能抽出多少个数, ...

  8. Educational Codeforces Round 37

    Educational Codeforces Round 37 这场有点炸,题目比较水,但只做了3题QAQ.还是实力不够啊! 写下题解算了--(写的比较粗糙,细节或者bug可以私聊2333) A. W ...

  9. Educational Codeforces Round 60 (Rated for Div. 2) - C. Magic Ship

    Problem   Educational Codeforces Round 60 (Rated for Div. 2) - C. Magic Ship Time Limit: 2000 mSec P ...

  10. Educational Codeforces Round 60 (Rated for Div. 2) - D. Magic Gems(动态规划+矩阵快速幂)

    Problem   Educational Codeforces Round 60 (Rated for Div. 2) - D. Magic Gems Time Limit: 3000 mSec P ...

随机推荐

  1. python爬虫实战(3)--图片下载器

    本篇目标 1.输入关键字能够根据关键字爬取百度图片 2.能够将图片保存到本地文件夹 1.URL的格式 进入百度图片搜索apple,这时显示的是瀑布流版本,我们选择传统翻页版本进行爬取.可以看到网址为: ...

  2. 关于TortoiseGit设置代理服务器会自动取消的问题解决方法

    在前些日子,我使用TortoiseGit的并给其设置代理服务器的时候,每次设置好之后,关闭打开总是会自动取消勾选,通过不断的尝试和分析之后,得出了解决方案. 首先,这个配置项是保存在HOME环境变量对 ...

  3. 详解CSS display:inline-block的应用(转)

    详解CSS display:inline-block的应用   阅读目录 基础知识 inline-block的问题 inline-block的应用 总结 本文详细描述了display:inline-b ...

  4. JS中的数组排序函数sort()

    JavaScript实现多维数组.对象数组排序,其实用的就是原生的sort()方法,用于对数组的元素进行排序. sort() 方法用于对数组的元素进行排序.语法如下: arrayObject.sort ...

  5. SpringBoot11 读取properties文件、发送邮件

    1 读取properties文件 1.1 ResourceBundle 帮助我们事先国际化 1.1.1 前提 properties文件的命名方式必须体现除语言和国别 例如:test_zh_CN.pro ...

  6. 最短路径Dijkstar算法和Floyd算法详解(c语言版)

    博客转载自:https://blog.csdn.net/crescent__moon/article/details/16986765 先说说Dijkstra吧,这种算法只能求单源最短路径,那么什么是 ...

  7. EZOJ #77

    传送门 分析 一个比较神奇的思路 我们考虑分治,对于每一个区间[le,ri]我们计算这个区间中左端点属于[le,mid],右端点属于[mid+1,ri]的情况对答案的贡献 我们求左半个区间的最大最小值 ...

  8. Django框架 之 URLconf

    Django框架 之 URLconf 浏览目录 URL 摘要 Django如何处理一个请求 反向解析URL name模式 namespace模式 一.URL 1.摘要 我们要在Django项目中为应用 ...

  9. java获取本机名称、IP、MAC地址和网卡名称

    java获取本机名称.IP.MAC地址和网卡名称 摘自:https://blog.csdn.net/Dai_Haijiao/article/details/80364370 2018年05月18日 1 ...

  10. Entity Framework Code-First(23):Entity Framework Power Tools

    Entity Framework Power Tools: Entity Framework Power Tools (currently in beta 3) has been released. ...