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.

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.

没错我超时了···     test19肯定是个大数据，然而一时想不出什么方法来优化。占坑留着吧    日后再补

 #include <stdio.h>

 #include <stdlib.h>

 int books[];

 int main(int argc, char *argv[])

 {

     int n,m;

     __int64 ways;

     while(scanf("%d%d",&n,&m)!=EOF)

     {

         ways=;

         int i,j;

         //for(i=0;i<n;i++)

             //scanf("%d",&books[i]);

         scanf("%d",&books[]);

         for(i=;i<n;i++)

         {

             scanf("%d",&books[i]);

             for(j=;j<i;j++)

             {

                 if(books[i]!=books[j])

                     ways++;

             }

         }

         printf("%I64d\n",ways);

     }

     return ;

 }

#include <stdio.h>

#include <stdlib.h>

#include <memory.h>

int a[],b[];//b组用来记录书的种类

int main(int argc, char *argv[])

{

    int n,m;

    int i,count;

    int k; 

    while(scanf("%d%d",&n,&m)!=EOF)

    {

        memset(b,,sizeof());

        k=;

        for(i=;i<n;i++)

        {

            scanf("%d",&a[i]);

            b[a[i]]++;    //相应种类的书籍数目++

        }

        count=n;

        for(i=;i<=m;i++)

        {

            count=count-b[i];

            k+=b[i]*count;//b[i]种类书的数量乘上剩下其他种类书籍的数量

        }

        printf("%d\n",k);

    }

    return ;

}