UESTC 1050 Different game 构造法

Different game

Time Limit: 3000/1000MS (Java/Others)     Memory Limit: 65535/65535KB (Java/Others)

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Alice is playing a new game recently. In this game, there are  n  different kinds of cards. We assume that Alice have  ci  pieces of cards for  ith  kind.

Alice is asked to divide them into  m m piles and then arrange each pile in one line. After that, Alice will get  m m sequences. For convenience, 

the sequences are labeled  S1,S2,⋯,Sm S1,S2,⋯,Sm. For each  i<j , Alice will get some points, equal to the length of 

the LCS(longest common subsequence) of  Si  and  Sj . The total points is the sum of points for all  i<j .

Now Alice is wondering the maximum points she can get.

As is known to everyone of you, Bob loves Alice very much. Could you tell Bob the answer to help Bob leave a good impression on Alice.

Input

The first line contains  2  integers  m  and  n , indicating the number of sequences and the number of different kinds of card.

The second line contains  n  integers  ci , indicating the number of  ith  card.

It is guaranteed that  1≤n,m≤100000,0≤ci≤100000      .

Output

Print the answer module 1000000007 in one line.

Sample input and output

Sample Input	Sample Output
2 2 2 3	2

Source

The 13th UESTC Programming Contest Final

My Solution

把第 i 种card的ci张卡先 ci/m分配给m个piles，多余的 ci % m 丢到最后，然后用 n*(n-1)/2  和 m*(m-1)/2 每次输入ci的时候计算并取模就好；

大致像这样构造

1 1 1 1 2 2 3 3 3 3 3 3 1 2 3

1 1 1 1 2 2 3 3 3 3 3 3 1 2 3

1 1 1 1 2 2 3 3 3 3 3 3 1 2 

1 1 1 1 2 2 3 3 3 3 3 3    2 

1 1 1 1 2 2 3 3 3 3 3 3    2 

1 1 1 1 2 2 3 3 3 3 3 3       

#include <iostream>
#include <cstdio>
using namespace std;
const int HASH = 1000000007;
int main()
{
    int m, n, c;
    long long ans = 0;
    scanf("%d%d", &m, &n);
    for(int i = 1; i <= n; i++){
        scanf("%d", &c);
        ans = (ans + (c/m)*(m*(m-1)/2) + (c%m)*(c%m-1)/2) % HASH;
    }
    printf("%lld", ans);
    return 0;
}

Thank you!