打印十字图

问题描述小明为某机构设计了一个十字型的徽标（并非红十字会啊），如下所示：..$$$$$$$$$$$$$....$...........$..$$$.$$$$$$$$$.$$$$...$.......$...$$.$$$.$$$$$.$$$.$$.$...$...$...$.$$.$.$$$.$.$$$.$.$$.$.$...$...$.$.$$.$.$.$$$$$.$.$.$$.$.$...$...$.$.$$.$.$$$.$.$$$.$.$$.$...$...$...$.$$.$$$.$$$$$.$$$.$$...$.......$...$$$$.$$$$$$$$$.$$$..$...........$....$$$$$$$$$$$$$..对方同时也需要在电脑dos窗口中以字符的形式输出该标志，并能任意控制层数。输入格式一个正整数 n (n<30) 表示要求打印图形的层数。输出格式对应包围层数的该标志。样例输入11样例输出1..$$$$$....$...$..$$$.$.$$$$...$...$$.$$$$$.$$...$...$$$$.$.$$$..$...$....$$$$$..样例输入23样例输出2..$$$$$$$$$$$$$....$...........$..$$$.$$$$$$$$$.$$$$...$.......$...$$.$$$.$$$$$.$$$.$$.$...$...$...$.$$.$.$$$.$.$$$.$.$$.$.$...$...$.$.$$.$.$.$$$$$.$.$.$$.$.$...$...$.$.$$.$.$$$.$.$$$.$.$$.$...$...$...$.$$.$$$.$$$$$.$$$.$$...$.......$...$$$$.$$$$$$$$$.$$$..$...........$....$$$$$$$$$$$$$..提示
请仔细观察样例，尤其要注意句点的数量和输出位置。
#include <iostream>#include <cstdio>#include <cstring>using namespace std;char c[333][333];int main(){    int n,i,j,t,k,m;    scanf("%d",&n);    k=4*n+5;    m=2*n+3;    for(i=1;i<=k;i++)        for(j=1;j<=k;j++)        c[i][j]='.';    for(i=m-2;i<=m+2;i++)//打印中心的        c[i][m]=c[m][i]='$';    t=1;    while(t<=n)    {        for(i=m-2*t;i<=m+2*t;i++)            c[i][m-2-2*t]=c[i][m+2+2*t]=c[m-2-2*t][i]=c[m+2+2*t][i]='$';        int l=m-2-2*t,r=m+2+2*t;        //四个角落        c[l][l]=c[l-1][l]=c[l][l-1]='$';        c[l][r]=c[l-1][r]=c[l][r+1]='$';        c[r][r]=c[r+1][r]=c[r][r+1]='$';        c[r][l]=c[r+1][l]=c[r][l-1]='$';        t++;    }    //把最外层的四个角还原    c[1][k]=c[k][1]=c[1][1]=c[k][k]='.';    for(i=1;i<=k;i++)    {        for(j=1;j<=k;j++)            printf("%c",c[i][j]);        cout<<endl;    }    return 0;}