计算和打印矩阵对角线的总和

Question

这个程序应该计算矩阵中所有对角线的总和，然后将它们打印出来。

例如。如果矩阵是

1 2 3 4 5
2 3 4 5 6
0 1 1 2 5
5 5 5 5 5
7 8 9 7 7

输出应该是

17 13 13 10 5

15 17 13 13 10

14 15 17 13 13

13 14 15 17 13

7 13 14 15 17

#include <stdio.h>

int main()
{
    int n, sum=0, i, j, sub_i, sub_j, sub1_i, sub1_j;

    scanf("%d ", &n);

    int array1[n][n];

    for(i=0;i<n;i++){
        for(j=0; j<n; j++){
            scanf("%d", &array1[i][j]);
        }
    }
    for(i=0; i<n; i++){
        for(j=0; j<n; j++){
            sub_i=i;
            sub_j=j;
            sub1_i=i;
            sub1_j=j;
            sum=0;

            if(j>i){
                while(sub_j<n){
                    sum+=array1[sub_i][sub_j];
                    sub_i++;
                    sub_j++;
                }
                while(sub_j<n){
                    array1[sub_i][sub_j]=sum;
                    sub1_i++;
                    sub1_j++;
                }
            }

            if(i>j){
                while(sub_i<n){
                    sum+=array1[sub1_i][sub1_j];
                    sub_i++;
                    sub_j++;
                }
                while(sub1_i<n){
                    array1[sub1_i][sub1_j]=sum;
                    sub1_i++;
                    sub1_j++;
                }
            }
        }
    }

    for(i=0; i<n; i++){
        for(j=0; j<n; j++){
            printf("%d ", array1[i][j]);
        }
        printf("\n");
    }
    return 0;
}

当我运行程序时，它会打印数组，就好像没有值分配给矩阵一样。有人可以指出发生了什么吗？

Answer 1

引用 Weather Vane 的评论:

The program alters the array it is examining — see array1[sub_i][sub_j]=sum; — and then prints incorrect values, since you can't correctly sum the diagonals of an array that is changing.

OP 已经意识到

... what you are telling me is to assign the values to another array and print that.

是的，这更容易:

#include <stdio.h>

int main(void)
{
    int n;
    // Checking the input is always a good idea, but you
    // may prefer something less brutal, in case of error
    if (scanf("%d", &n) != 1  ||  n < 1)
       return 1;    

    int mat[n][n];

    for (int i = 0; i < n; ++i) {
        for (int j= 0; j < n; ++j) {
            if (scanf("%d", &mat[i][j]) != 1)
                return 1; 
        }
    }

    // Calculate and store the sum of the diagonals. Note that
    // it could be done in the previous loop, but it may be better
    // to refactor those snippets into separate functions.
    int diagonals[2 * n + 1];
    for (int i = 0; i < 2 * n + 1; ++i)
        diagonals[i] = 0;  // consider 'memset' instead of this loop

    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < n; ++j) {
            diagonals[n + i - j] += mat[i][j]; 
        }    
    }

    // Now print the correct values in their position
    for(int i = 0; i < n; ++i) {
        for(int j = 0; j < n; ++j) {
            printf("%4d", diagonals[n + i - j]);
        }
        printf("\n");
    }
    return 0;
}

可测试 HERE .