Problem B: C语言习题 矩阵元素变换
Time Limit: 1 Sec Memory Limit: 128 MBSubmit: 942 Solved: 558
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Description
函数实现。用main函数调用。
Input
输入n和矩阵中的每个元素
Output
变换后的矩阵
Sample Input
5
25 13 9 5 1
16 17 18 19 6
15 24 4 20 7
14 23 22 21 8
2 12 11 10 3 Sample Output
1 13 9 5 2
16 17 18 19 6
15 24 25 20 7
14 23 22 21 8
3 12 11 10 4 HINT
主函数已给定如下，提交时不需要包含下述主函数
/* C代码 */
int main()
{
void change(int *,int );
int **a,*p,i,j;
int n;
scanf("%d",&n);
p=(int*)malloc(n*n*sizeof(int));
a=(int**)malloc(n*sizeof(int *));
for(i=0; i<n; i++)
a[i]=p+n*i;
for (i=0; i<n; i++) //输入矩阵
for (j=0; j<n; j++)
scanf("%d",&a[i][j]);
change(p,n); //调用函数，实现交换
for (i=0; i<n; i++) //输出已交换的矩阵
{
for (j=0; j<n; j++)
printf("%d ",a[i][j]);
printf("\n");
}
free(p);
free(a);
return 0;
}
/* C++代码 */
int main()
{
void change(int *,int );
int **a,*p,i,j;
int n;
cin>>n;
p=new int[n*n];
a=new int*[n];
for(i=0; i<n; i++)
a[i]=p+n*i;
for (i=0; i<n; i++) //输入矩阵
for (j=0; j<n; j++)
cin>>a[i][j];
change(p,n); //调用函数，实现交换
for (i=0; i<n; i++) //输出已交换的矩阵
{
for (j=0; j<n; j++)
cout<<a[i][j]<<" ";
cout<<endl;
}
delete []p;
delete []a;
return 0;
}
#include<stdio.h>
#include<stdlib.h>
void change(int *p,int n)
{int max=0,maxj,m1=99,m2=99,m3=99,min=99,minj,m1j,m2j,m3j,x;int t;for (x=0; x<n*n; x++)if(*(p+x)>max){max=*(p+x);maxj=x;}t=*(p+maxj);*(p+maxj)=*(p+(n*n)/2);*(p+(n*n)/2)=t;for (x=0; x<n*n; x++)if(*(p+x)<min){min=*(p+x);minj=x;}t=*(p+minj);*(p+minj)=*(p+0);*(p+0)=t;for (x=0; x<n*n; x++)if(*(p+x)>min&&*(p+x)<m1){m1=*(p+x);m1j=x;}t=*(p+m1j);*(p+m1j)=*(p+(n-1));*(p+(n-1))=t;for (x=0; x<n*n; x++)if(*(p+x)>m1&&*(p+x)<m2){m2=*(p+x);m2j=x;}t=*(p+m2j);*(p+m2j)=*(p+n*(n-1));*(p+n*(n-1))=t;for (x=0; x<n*n; x++)if(*(p+x)>m2&&*(p+x)<m3){m3=*(p+x);m3j=x;}t=*(p+m3j);*(p+m3j)=*(p+n*n-1);*(p+n*n-1)=t;}
int main()
{void change(int *,int );int **a,*p,i,j;int n;scanf("%d",&n);p=(int*)malloc(n*n*sizeof(int));a=(int**)malloc(n*sizeof(int *));for(i=0; i<n; i++)a[i]=p+n*i;for (i=0; i<n; i++)for (j=0; j<n; j++)scanf("%d",&a[i][j]);change(p,n);for (i=0; i<n; i++){for (j=0; j<n; j++)printf("%d ",a[i][j]);printf("\n");}free(p);free(a);return 0;
}
)
)
