Educational Codeforces Round 3 B
2 seconds
256 megabytes
standard input
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.
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.
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.
4 3
2 1 3 1
5
7 4
4 2 3 1 2 4 3
18
The answer to the first test sample equals 5 as Sasha can choose:
- the first and second books,
- the first and third books,
- the first and fourth books,
- the second and third books,
- 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
- [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 ...
- [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 ...
- [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 ...
- [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 ...
- [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 ...
- 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 ...
- 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, ...
- 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 ...
- 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 ...
- 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
- hibernate学习笔记(4)表单操作
User.hbm.xml的表单配置: ①主键 <id name="id" type="java.lang.Integer"> <column ...
- c语言中argc和argv
main函数的参数,解释如下: argc:命令行总的参数的个数,即argv中元素的格式. *argv[ ]:字符串数组,用来存放指向你的字符串参数的指针数组,每一个元素指向一个参数. argv[0]: ...
- php apc 安装
APC简介 APC(Alternative PHP Cache)是一个PHP缓存.它在内存中存储PHP页面并且减少了硬盘的I/O.这对于性能的提升十分明显.你甚至可以在CPU使用率下降50%的情况下提 ...
- cocos2dx中的内存管理方式
转载:http://www.cocoachina.com/bbs/read.php?tid=195219 今天看了一下cocos2dx的内存管理机制,有些地方不太好理解搞了挺长的时间,现在感觉自己理解 ...
- Windows libQGLViewer2.7.0,libQGLViewer2.6.2与g2o20160427, g2o20170730编译生成G2O
1. Win10, VS2013, libQGLViewer2.6.2 和 QT5.6.3编译时候会出错,初步判断libQGLViewer2.6.2不支持QT5 错误 error LNK1120: 个 ...
- hdu6331 Problem M. Walking Plan
传送门 题目大意 给你一个n点m条边的有向图,q次询问,给定s,t,k,求由s到t至少经过k条边的最短路. 分析 我们设dp[i][j][k]为从i到j至少经过k条边的最短路,sp[i][j]意为从i ...
- 100741A Queries
传送门 题目 Mathematicians are interesting (sometimes, I would say, even crazy) people. For example, my f ...
- Luogu 1314 [NOIP2011] 聪明的质监员
二分答案 + 前缀和. 题面中式子的意思是每一个区间$[l, r]$的贡献是这个区间内$w_i \geq W$的个数乘以这些$i$的$v_i$和. 很快发现了答案具有单调性,可以做两遍二分,分别看看小 ...
- Servlet入门第二天
1. GenericServlet: 1). 是一个 Serlvet. 是 Servlet 接口和 ServletConfig 接口的实现类. 但是一个抽象类. 其中的 service 方法为抽象方法 ...
- 数据结构 lucky_ming幸运的小明
问题描述 在快速排序过程中, 每次会找一个划分值, 将小于划分值的放到其左边, 大于划分值的放右边, 然后再依次递归左右两边, 对子序列进行同样的操作, 直到子序列为空则停止操作.最后就得到了有序的序 ...