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Copy path数字旋转方阵.cpp
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数字旋转方阵.cpp
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59 lines (58 loc) · 1.8 KB
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#include<bits/stdc++.h>
using namespace std;
int data[100][100];
void Full(int number, int begin, int size);
int main()
{
int size;
cout<<"输入方阵的大小:";
cin>>size;
Full(1,0,size);
for(int i=0; i<size; i++)
{
for(int j=0; j<size; j++)
cout<<data[i][j]<<setw(4);
cout<<endl;
}
cout<<endl;
return 0;
}
void Full(int number, int begin, int size)
{
//从number 开始填写size 阶方阵,左上角的下标为(begin, begin)
int i, j, k;
if (size == 0) //递归的边界条件,如果size等于0,则无须填写
return;
if (size == 1) //递归的边界条件,如果size等于1
{
data[begin][begin] = number; //则只须填写number
return;
}
i = begin;
j = begin; //初始化左上角下标
for (k = 0; k < size - 1; k++) //填写区域A,共填写size - 1个数
{
data[i][j] = number; //在当前位置填写number
number++;
i++; //行下标加1
}
for (k = 0; k < size - 1; k++) //填写区域B,共填写size - 1个数
{
data[i][j] = number; //在当前位置填写number
number++;
j++; //列下标加1
}
for (k = 0; k < size - 1; k++) //填写区域C,共填写size - 1个数
{
data[i][j] = number; //在当前位置填写number
number++;
i--; //行下标减1
}
for (k = 0; k < size - 1; k++) //填写区域D,共填写size - 1个数
{
data[i][j] = number; //在当前位置填写number
number++;
j--; //列下标减1
}
Full(number, begin + 1, size - 2); //递归求解,左上角下标为begin + 1
}