codeforces 691E（矩阵乘法）

最新推荐文章于 2024-06-27 15:51:59 发布

转载最新推荐文章于 2024-06-27 15:51:59 发布 · 155 阅读

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原文链接：http://www.cnblogs.com/hnqw1214/p/6498590.html

本文介绍了一种使用矩阵乘法解决Xor-序列计数问题的方法。通过预处理矩阵来判断两个数异或后的结果是否能被3整除，并通过矩阵的幂运算求解特定长度的Xor-序列数量。

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You are given n integers a1, a2, ..., an.

A sequence of integers x1, x2, ..., xk is called a "xor-sequence" if for every 1 ≤ i ≤ k - 1 the number of ones in the binary representation of the number xi $\text{[math]}$ xi + 1's is a multiple of 3 and $\text{[math]}$ for all 1 ≤ i ≤ k. The symbol $\text{[math]}$ is used for the binary exclusive or operation.

How many "xor-sequences" of length k exist? Output the answer modulo 109 + 7.

Note if a = [1, 1] and k = 1 then the answer is 2, because you should consider the ones from a as different.

Input

The first line contains two integers n and k (1 ≤ n ≤ 100, 1 ≤ k ≤ 1018) — the number of given integers and the length of the "xor-sequences".

The second line contains n integers ai (0 ≤ ai ≤ 1018).

Output

Print the only integer c — the number of "xor-sequences" of length k modulo 109 + 7.

Examples

input

5 2
15 1 2 4 8

output

input

5 1
15 1 2 4 8

output

5
思路：很好的一道矩阵乘法题，需要建立起模型。先预处理出一个矩阵，第i行j个表示a[i]异或a[j]是否能被3整除，然后再将矩阵计算k-1次幂，最后将矩阵中所有元素加起来。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<cstring>
#include<string>
#include<vector>
#include<map>
#include<set>
#include<queue>
#define mo 1000000007
using namespace std;
struct Matrix
{
    long long a[101][101];
};
long long n,k,a[101];
Matrix st,ans;
Matrix mul(Matrix a,Matrix b)
{
    Matrix c;
    int i,j,k;
    //<F5>a.a[1][1]=1;
    for (i=1;i<=n;i++)
        for (j=1;j<=n;j++)
        {
            c.a[i][j]=0;
            for (k=1;k<=n;k++)
                c.a[i][j]=(c.a[i][j]+(a.a[i][k]*b.a[k][j]%mo))%mo;
        }
    return c;
}
Matrix power(Matrix a,long long b)
{
    Matrix c;
    int i;
    memset(c.a,0,sizeof(c.a));
    for (i=1;i<=n;i++) c.a[i][i]=1;
    while (b)
    {
        if (b&1) c=mul(c,a);
        b>>=1;
        a=mul(a,a);
    }
    return c;
}
bool can(long long x)
{
    long long ret=0;
    while (x)
    {
        ret+=(x&1);
        x>>=1;
    }
    return (ret%3==0);
}
int main()
{
    scanf("%lld%lld",&n,&k);
    int i,j;
    for (i=1;i<=n;i++) scanf("%lld",&a[i]);
    for (i=1;i<=n;i++)
        for (j=1;j<=n;j++) st.a[i][j]=can(a[i]^a[j]);
    ans=power(st,k-1);
    long long ret=0;
    //cout<<ans.a[i][j]<<endl;
    for (i=1;i<=n;i++)
        for (j=1;j<=n;j++) ret=(ret+ans.a[i][j])%mo;
    printf("%lld\n",ret);
    return 0;
}

转载于:https://www.cnblogs.com/hnqw1214/p/6498590.html