牛客多校第八场 C CDMA 线性代数：沃尔什矩阵

题意：

构造出一个由1和-1组成的$2^k*2^k$的矩阵，使得矩阵任意两列内积为0

题解：

数学知识，沃尔什矩阵。沃尔什矩阵的特性被CDMA（码分多址）采用，使得编码成为无线信号的频段和振幅之外的第三维，提高了无线信道利用率。

构造沃尔什矩阵只需倍增构造，以第i个矩阵的第k行重复两遍，作为第i+1个矩阵的第2k行，取正一遍，取反一遍，作为第i+1个矩阵的第2k+1行。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<string>
#include<stack>
#include<algorithm>
#include<map>
#include<queue>
#include<vector>
using namespace std;
#define INF 0x3f3f3f3f
#define MAXN 100000+50
#define MAXM 30000
#define ll long long
#define per(i,n,m) for(int i=n;i>=m;--i)
#define rep(i,n,m) for(int i=n;i<=m;++i)
#define mod 1000000000 + 7
#define mian main
#define mem(a, b) memset(a, b, sizeof a)
#ifndef ONLINE_JUDGE
#define dbg(x) cout << #x << "=" << x << endl;
#else
#define dbg(x)
#endif
inline int read()
{
    int x = , f = ;
    char ch = getchar();
    while (ch < '' || ch > '')
    {
        if (ch == '-')
            f = -;
        ch = getchar();
    }
    while (ch >= '' && ch <= '')
    {
        x =  * x + ch - '';
        ch = getchar();
    }
    return x * f;
}
inline ll readll()
{
    ll x = , f = ;
    char ch = getchar();
    while (ch < '' || ch > '')
    {
        if (ch == '-')
            f = -;
        ch = getchar();
    }
    while (ch >= '' && ch <= '')
    {
        x =  * x + ch - '';
        ch = getchar();
    }
    return x * f;
}
int a[][][];
void init()
{
    a[][][] = ;
    for (int i = ; i <= ; ++i)
    {
        int t = pow(, i-);
        for (int j = ; j <= t;j++)
        {
            for (int k = ; k <= t / ; ++k)
            {
                a[i][j][k] = a[i - ][(j + ) / ][k];
            }
            if (j & )
            {
                for (int k = t/+; k <= t; ++k)
                {
                    a[i][j][k] = a[i - ][(j + ) / ][k-t/];
                }
            }
            else
            {
                for (int k = t /  + ; k <= t; ++k)
                {
                    a[i][j][k] =- a[i - ][(j + ) / ][k-t/];
                }
            }
        }
    }
}
int main()
{
    int _ = ;
    init();
    while (_--)
    {
        int n = read();
        int num = ;
        int t = n;
        while (t)
        {
            t >>= ;
            num++;
        }
        for(int i=;i<=n;++i)
            for (int j = ; j <= n; ++j)
            {
                printf("%d%c", a[num][i][j], " \n"[j == n]);
            }
    }
}