The basic algorithm of C language 35-the product of the product pair diagonal of the product of the triangle on the array

Array algorithm
/*
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Title: The product of the triangle on the four-order matrix, the product of the main diagonal, the product of the Vice-diagonal:
Such as:
8 3 6 5
0 4 3 2
0 6 1 5
7 0 0 2
The upper triangular product is: 172800
The main diagonal product is: 64
The product of negative diagonal is: 630
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*/
#include <stdio.h>
void Main ()
{
int i,j;
int ssj=1,zdj=1,fdj=1;
int a[4][4];
printf ("Input 4*4 matrix: \ n");
for (i=0;i<4;i++)
for (j=0;j<4;j++)
scanf ("%d", &a[i][j]);
printf ("Output matrix: \ n");
for (i=0;i<4;i++)
{
for (j=0;j<4;j++)
printf ("%3d", A[i][j]);
printf ("\ n");
}
printf ("Output of the triangle on the product:");
for (i=0;i<4;i++)
for (j=i;j<4;j++)
SSJ*=A[I][J];
printf ("%d\n", SSJ);

printf ("Product of the main diagonal:");
for (i=0;i<4;i++)
Zdj*=a[i][i];
printf ("%d\n", ZDJ);

printf ("Product of the Diagonal Line:");
for (i=3;i>=0;i--)
Fdj*=a[i][3-i];
printf ("%d\n", FDJ);
}
/*
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Evaluation:

The upper triangle is the triangle above the main diagonal, the condition should be: line from 0 to n-1, column from corresponding line number to n-1;

(i=0;i<=n-1;i++) (j=i;j<=n-1;j++); The main diagonal is the connection from a[0][0] to a[n-1][n-1], the condition is:

Rows from 0 to n-1, column = row, (i=0;i<n-1;i++) corresponds to a product of a[i][i]; In the same vein, the negative diagonal is from a[n-1][0] to

A[0][n-1], the condition is: line from n-1 to 0, column =n-1-row, (i=n-1;i>=0;i--), product item is

A[i][n-1-i]; it's easy to solve the problem by analyzing it! In fact, this algorithm can be extended to any order matrix, read

Who are interested in completing their own!

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*/

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The basic algorithm of C language 35-the product of the product pair diagonal of the product of the triangle on the array