黑白棋子的移动 简单递归一下下就好了 嘿嘿

解题思路:

                                                                                                除了后4行以外

                                                                                                每两行分一组的时候,可以很明显地看到规律.

                                                                                                中间的 "o*" 与 "--" 交换

                                                                                                最右边的 "**" 与 "--" 交换

                                                                                                而后4行则按照样例的交换步骤即可

                                                                                                
注意事项:

参考代码:

#include"bits/stdc++.h"
using namespace std;
char a[300];
int n,kkk;
void print(){//输出当前状态
    printf("step%2d:",kkk);
    for(int i=1;i<=2*n+2;i++) cout<<a[i];
    cout<<endl;
    kkk++;
}
void f(int k){
    swap(a[k],a[k/2]);
    swap(a[k+1],a[k/2+1]);
    print();
    if(k/2<=4) return;//递归结束条件，及后4行特殊处理
    swap(a[k-2],a[k/2]);
    swap(a[k-1],a[k/2+1]);

    print();
    f(k-2);
}
int main(){
    cin>>n;
    kkk=0;
    for(int i=1;i<=2*n+2;i++){
        if(i<=n) a[i]='o';
        else if(i>n&&i<=2*n) a[i]='*';
        else a[i]='-';
    }
    print();
    f(2*n+1);
    //以下是处理后4行特殊情况
    swap(a[4],a[8]);
    swap(a[5],a[9]);
    print();
    swap(a[2],a[8]);
    swap(a[3],a[9]);
    print();
    swap(a[2],a[7]);
    swap(a[3],a[8]);
    print();
    swap(a[1],a[7]);
    swap(a[2],a[8]);
    print();
    return 0;
}