juzhen.c

设计算法求三对角矩阵在压缩存储下的转置矩阵。

#include<stdio.h>
#include<malloc.h>
#define max 100
typedef struct node{
    int n;
    int data[max];
}duijiao;

duijiao *Set()
{
    duijiao *A;
    int i;
    A=(duijiao *)malloc(sizeof(struct node));
    printf("Please enter a tridiagonal matrix of order:");
    scanf("%d",&A->n);
	printf("Please enter the tridiagonal matrix of all non-zero elements of <% d months>:\n",3*(A->n)-2);
	for(i=0;i<3*(A->n)-2;i++)
    {
        scanf("%d",&A->data[i]);
    }
    printf("Tridiagonal matrix of elements from top to bottom, are arranged in chronological order from left to right:\n");
    return A;
}

void zhuanzhi(duijiao *A)
{
    int i=0;
    int t;
	while(i<3*(A->n)-2)
    {
        if(i%3==0||i==0)
          i++;
        else
        {
            t=A->data[i];
            A->data[i]=A->data[i+1];
            A->data[i+1]=t;
            i+=2;
        }
    }
}

void print(duijiao *A)
{
    int i;
	for(i=0;i<3*(A->n)-2;i++)
    {
        printf("%d ",A->data[i]);
    }
    printf("\n");
}

void result(duijiao *A)
{
    int i,j,k,m;
    int a[50][50];
    for(i=0;i<(A->n);i++)
    {
        switch(i)
        {
            case 0:
                   for(j=0;j<2;j++)
                   {
                     a[i][j]=A->data[2*i+j];
                   }
                   for(j=2;j<(A->n);j++)
                   {
                     a[i][j]=0;
                   }
                   break;
            
           case 4:
                   for(j=(A->n)-2;j<(A->n);j++)
                   {
                     a[i][j]=A->data[2*i+j];
                   }
                   for(j=0;j<(A->n)-2;j++)
                   {
                     a[i][j]=0;
                   }
                   break;
          default:
                   for(j=i-1;j<i-1+3;j++)
                   {
                    a[i][j]=A->data[2*i+j];
                   }
                   for(j=0;j<i-1;j++)
                   {
                    a[i][j]=0;
                   }
                   for(j=i-1+3;j<(A->n);j++)
                   {
                    a[i][j]=0;
                   }
                   break;
        }
    }
	printf("Results:\n");
    for(k=0;k<(A->n);k++)
    {
        for(m=0;m<(A->n);m++)
        {
           printf("%3d ",a[k][m]);
        }
        printf("\n");
    }
}

void main()
{
    duijiao *A;
    int x;
    loop:
    {
    	A=Set();
        zhuanzhi(A);
        print(A);
        result(A);
    }
    printf("Do you want to continue to operate?YES(1)-NO(2):");
    scanf("%d",&x);
    switch(x)
	{
	case 1:
	    goto loop;
	    break;
    case 2:
		break;
	}
}