鏈接：https://ac.nowcoder.com/acm/contest/888/C
來源：牛客網

Gromah and LZR have entered the third level. There is a blank grid of size m\times mm×m, and above the grid is a word “CDMA”.

In CDMA Technology, a Technology about computer network, every network node should be appointed a unique binary sequence with a fixed and the same length. Moreover, if we regard 0_{}0
?
in the sequence as -1_{}?1
?
, and regard 1_{}1
?
as +1_{}+1
?
, then these sequences should satisfy that for any two different sequences s,t_{}s,t
?
, the inner product of s,t_{}s,t
?
should be exactly 0_{}0
?
.

The inner product of two sequences s,t_{}s,t
?
with the same length n_{}n
?
equals to \sum_{i=1}^{n} s_it_i∑

So, the key to the next level is to construct a grid of size m\times mm×m, whose elements are all -1_{}?1
?
or 1_{}1
?
, and for any two different rows, the inner product of them should be exactly 0_{}0
?
.

In this problem, it is guaranteed that m_{}m
?
is a positive integer power of 2_{}2
?
and there exists some solutions for input m_{}m
?
. If there are multiple solutions, you may print any of them.
輸入描述:
Only one positive integer m_{}m
?
in one line.

m \in {2^k ; | ;k = 1, 2, \cdots, 10}m∈{2
k
∣k=1,2,?,10}
輸出描述:
Print m_{}m
?
lines, each contains a sequence with length m_{}m
?
, whose elements should be all -1_{}?1
?
or 1_{}1
?
satisfying that for any two different rows, the inner product of them equals 0_{}0
?
.

You may print multiple blank spaces between two numbers or at the end of each line, and you may print any number of blank characters at the end of whole output file.
示例1
輸入
復制
2
輸出
復制
1 1
1 -1
說明
The inner product of the two rows is 1\times(-1) + 1\times 1 = 01×(?1)+1×1=0.
構造1，-1矩陣使得每一行對應位置的數乘積加起來都為零。
典型的構造題，不禁想起來課本上的循環賽程表，根據小的去構造大的，通過給出的m=2的去構造m=4的，然后找出規律后構造出m=8的發現也對，就寫了。代碼中解釋
代碼如下：

#include<bits/stdc++.h> #define ll long long using namespace std;int a[1030][1030]; int n;int main() {while(scanf("%d",&n)!=EOF){a[1][1]=1;a[1][2]=1;a[2][1]=-1;a[2][2]=1;if(n==2){for(int i=1;i<=2;i++){for(int j=1;j<=2;j++)printf("%d ",a[i][j]);printf("\n");}}else{int y=4;while(y<=n){for(int i=y/2+1;i<=y;i++)//左下方的矩陣就等于左上方的矩陣的相反數{for(int j=1;j<=y/2;j++)a[i][j]=-a[i-y/2][j];}for(int i=1;i<=y/4;i++)//右上方的矩陣前兩行左下方的矩陣前兩行{for(int j=y/2+1;j<=y;j++){a[i][j]=a[y/2+i-1][j-y/2];}}for(int i=y/4+1;i<=y/2;i++)//右上方的矩陣后兩行等于左下方的矩陣后兩行的相反數{for(int j=y/2+1;j<=y;j++){a[i][j]=-a[y/2+i-1][j-y/2];}}for(int i=y/2+1;i<=y;i++)//右下方的矩陣就等于右上方的矩陣{for(int j=y/2+1;j<=y;j++)a[i][j]=a[i-y/2][j];}y*=2;}for(int i=1;i<=n;i++){for(int j=1;j<=n;j++){printf("%d ",a[i][j]);}printf("\n");}}} }

努力加油a啊，(^o)/~

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