codeforces 691E(矩阵乘法)

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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E. Xor-sequences
time limit per test
3 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

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  xi  +  1's is a multiple of 3 and  for all 1 ≤ i ≤ k. The symbol  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
13
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

### Codeforces 887E Problem Solution and Discussion The problem **887E - The Great Game** on Codeforces involves a strategic game between two players who take turns to perform operations under specific rules. To tackle this challenge effectively, understanding both dynamic programming (DP) techniques and bitwise manipulation is crucial. #### Dynamic Programming Approach One effective method to approach this problem utilizes DP with memoization. By defining `dp[i][j]` as the optimal result when starting from state `(i,j)` where `i` represents current position and `j` indicates some status flag related to previous moves: ```cpp #include <bits/stdc++.h> using namespace std; const int MAXN = ...; // Define based on constraints int dp[MAXN][2]; // Function to calculate minimum steps using top-down DP int minSteps(int pos, bool prevMoveType) { if (pos >= N) return 0; if (dp[pos][prevMoveType] != -1) return dp[pos][prevMoveType]; int res = INT_MAX; // Try all possible next positions and update 'res' for (...) { /* Logic here */ } dp[pos][prevMoveType] = res; return res; } ``` This code snippet outlines how one might structure a solution involving recursive calls combined with caching results through an array named `dp`. #### Bitwise Operations Insight Another critical aspect lies within efficiently handling large integers via bitwise operators instead of arithmetic ones whenever applicable. This optimization can significantly reduce computation time especially given tight limits often found in competitive coding challenges like those hosted by platforms such as Codeforces[^1]. For detailed discussions about similar problems or more insights into solving strategies specifically tailored towards contest preparation, visiting forums dedicated to algorithmic contests would be beneficial. Websites associated directly with Codeforces offer rich resources including editorials written after each round which provide comprehensive explanations alongside alternative approaches taken by successful contestants during live events. --related questions-- 1. What are common pitfalls encountered while implementing dynamic programming solutions? 2. How does bit manipulation improve performance in algorithms dealing with integer values? 3. Can you recommend any online communities focused on discussing competitive programming tactics? 4. Are there particular patterns that frequently appear across different levels of difficulty within Codeforces contests?
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