基于c语言和java的螺旋方阵.doc

编程序，显示螺旋方阵：1234 1213145 1116156 10987#include#define max 10int main() int squaremaxmax;int i,j,k=1,d=0,n;/d为操作向量printf(请输入螺旋方阵的阶数n：);scanf(%d,&n);if(nmax)printf(阶数超过最大值max);printf(n);printf(Please try again); printf(n);elsefor(i=0;in;+i)for(j=0;jn;+j)squareij=0; /*将方阵各个元素都赋值为零*/ i=j=0;for(;k=n*n;k+) squareij=k;/给数组赋值switch(d) case 0: if(0=i&in&0=(j+1)&(j+1)n&squarei(j+1)=0) j+;/*实现列递加*/ else d=1,i+; break;/退出switch语句case 1:if(0=(i+1)&(i+1)n&0=j&jn&square(i+1)j=0) i+;/*实现行递加*/ else d=2,j-; break;case 2:if(0=i&in&0=(j-1)&(j-1)n&squarei(j-1)=0) j-;/*实现列递减*/ else d=3,i-; break;case 3:if(0=(i-1)&(i-1)n&0=j&jn&square(i-1)j=0) i-;/*实现行递减*/ else d=0,j+; break;default:break;for(int a=0;an;a+)for(int b=0;b20)System.out.print(阶数超过最大值);System.out.print(Please try again);elsefor(i=0;in;+i)for(j=0;jn;+j)squareij=0; /*将方阵各个元素都赋值为零*/ i=j=0;for(;k=n*n;k+) squareij=k;/给数组赋值switch(d) case 0: if(0=i&in&0=(j+1)&(j+1)n&squarei(j+1)=0) j+;/*实现列递加*/ else d=1; i+; break;/退出switch语句case 1:if(0=(i+1)&(i+1)n&0=j&jn&square(i+1)j=0) i+;/*实现行递加*/ else d=2; j-; break;case 2:if(0=i&in&0=(j-1)&(j-1)n&squarei(j-1)=0) j-;/*实现列递减*/ else d=3; i-; break;case 3:if(0=(i-1)&(i-1)n&0=j&jn&square(i-1)j=0) i-;/*实现行递减*/ else d=0; j+; break;default:break;for(i=0;in;i+)for(j=0;jn;j+)Sys