对称矩阵
时间复杂度
o(POW(N,2))
C++ 代码
#include<iostream>
#include<cstring>
using namespace std;
int main()
{
int a, d[101][101];
while (1)
{
memset(d, 0, sizeof d);
cin >> a;
if (a)
{
for (int i = 1; i <= a; i++)
{
for (int j = 1; j <= a; j++)
{
if (j < i)
cout << d[i][j] << " ";
else
{
d[i][j] = j - i+1;
d[j][i] = d[i][j];
cout << d[i][j] << " ";
}
}
cout << endl;
}
}
else break;
cout << endl;
}
}