Codeforces Tutorial
C. Harmony Analysis
Problem Analysis
题目大意生成一个维度为2的k次方的方阵，使得任意两行的内积为0.
当(k=2)时
[left[
egin{array}{cc|cc}
1 & 1 & 1 & 1 \
1 & -1 & 1 & -1 \
hline
1 & 1 & -1 & -1\
1 & -1 & -1 & 1\
end{array}
ight]
]
可以发现规律是除了右下角，其他三个矩阵相同，右下角每个元素去相反数。即
[left[
egin{array}{c|c}
A& A\
hline
A& -A\
end{array}
ight]
]
可以验证当(k=1)时也成立。
Acepted Code
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<string>
#include<vector>
#include<cmath>
#include<map>
#include<istream>
#include<cassert>
#include<set>
#define DEBUG(x) cout<<#x<<" = "<<x<<endl
#define DEBUG2(x,y) cout<<#x<<" = "<<x<<" , "
<<#y<<" = "<<y<<endl
using namespace std;
typedef long long ll;
const int MAXN=600;
int power2(int n)
{
int rt=1;
for(int ii=1;ii<=n ;ii++ ){
rt*=2;
}
return rt;
}
int grid[MAXN][MAXN];
void format(int n)
{
if(n==1)cout<<"+";
else cout<<"*";
}
int main()
{
// freopen("in.txt","r",stdin);
int k;
cin>>k;
grid[1][1]=1;
for(int ii=1;ii<=k ;ii++ ){
///对称生长法
int l=power2(ii-1)+1,r=power2(ii),
delta=l-1;
for(int row=l-delta;row<=r-delta ;row++ ){
for(int col=l;col<=r ;col++ ){
grid[row][col]=grid[row][col-delta];
}
}
for(int row=l;row<=r ;row++ ){
for(int col=l-delta;col<=r-delta ;col++ ){
grid[row][col]=grid[row-delta][col];
}
}
for(int row=l;row<=r ;row++ ){
for(int col=l;col<=r ;col++ ){
grid[row][col]=-grid[row-delta][col-delta];
}
}
}
int pw=power2(k);
for(int ii=1;ii<=pw ;ii++ ){
for(int jj=1;jj<=pw ;jj++ ){
format(grid[ii][jj]);
}
cout<<endl;
}
}
Wrong Answer Cases
What I Learn
给出的矩阵是(n imes n)，然后(n)又是(2)的幂次。突破口往往在这样比较凑巧的特点上。所以题目的形式值得反复推敲