DNA （C语言代码）

注意事项： 表达得非常乱，代码也很乱，就不贴了；

解题思路:  应用for循环；解决题目要求；

for( z=0; z<b; z++)  // 重复数嵌套行数

{

     int c; //判断当前是否是中间行 

     int d=0; // *号前的空格和 ，每次输出一行加 s，到

                                                                                X   X

                                                                                 X X

                                                                                  X   这里的时候，s 就                                                                                                                                                             s=-s;

                                                                                                             d += s;

                                                                                                             l = -l;

                                                                                                        就d就

                                                                                 X X

                                                                                X   X  

     int s=1; //空格数 

     int k = a;// 方便计算两个*号间的间隔   

     int l = 2; // 下边的*号 

  X

 X X

X   X

     int v =1;

    for( i=0; i<a; i++)

    {

         int j;

         c = k-l; // *  *  中间的空格个数等于 总行数-2 

         k -= l;  

         使用if判断，c是否大于0；是 则再判断v是否输出

                                                                                X   X

                                                                                 X X

                                                                                  X

                                                                                 X X

                                                                                X   X  这行

                                                                                 X X

                                                                                  X

                                                                                 X X

                                                                                X   X

        因为重复数。

for( j = d; j>0;j--)

{

printf(" ");            输出空格 X     X  

}

    }

}

注意事项:  输入判断：

while( !(N<=15))

{

scanf("%d",&N);

}

for( h=0; h<N; h++ )

{

scanf("%d%d",&a,&b);

while(!(a>=3 && a<=39 && a%2!=0) || !(b>=1 && b<=20)) //判断a,b是否符合要求  ！取非 

{

scanf("%d%d",&a,&b);

}

参考代码: