Problem B: C語言習題 矩陣元素變換
Time Limit: 1 Sec??Memory Limit: 128 MBSubmit: 942??Solved: 558
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Description
函數實現。用main函數調用。
Input
輸入n和矩陣中的每個元素
Output
變換后的矩陣
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
主函數已給定如下，提交時不需要包含下述主函數
/* C代碼 */
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++)???????????????????? //輸入矩陣
??????? for (j=0; j<n; j++)
??????????? scanf("%d",&a[i][j]);
??? change(p,n);????????????????????????????? //調用函數，實現交換
??? for (i=0; i<n; i++)??????????????????? //輸出已交換的矩陣
??? {
??????? for (j=0; j<n; j++)
??????????? printf("%d ",a[i][j]);
??????? printf("\n");
??? }
??? free(p);
??? free(a);
??? return 0;
}
/* C++代碼 */
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++)???????????????????? //輸入矩陣
??????? for (j=0; j<n; j++)
??????????? cin>>a[i][j];
??? change(p,n);?????????????????????????? //調用函數，實現交換
??? for (i=0; i<n; i++)??????????????????? //輸出已交換的矩陣
??? {
??????? 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;
}
)
)
