求n阶方阵中各条反斜线上的元素之和4*4

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#include<stdio.h>void main(){//其中第一条斜线是00 - 11 - 22 -33 第二条10 - 21 - 32int arr2[4][4] = { 00,  01,   02, 03,  10 ,  11,   12, 13,  20 ,  21,   22, 23,  30,   31,   32, 33,};int i, j;int sum = 0;int index = 0;for (int i = 0; i < 4; i++){for (int j = 0; j < 4; j++){//printf("最初的i=%d", i);//满足这个条件的情况下if (j - i == j){int index = j ;printf("\n\n\n%d\n\n", index);printf("i=%d,j=%d\n", i, j);for (int i = 0; i < 4; i++){for (int j = 0; j < 4; j++){if (j - i == index){sum += arr2[i][j];}}}printf("y行的数                 据时%d\n", sum);system("pause");sum = 0;}if (i - j == i){int index = i;printf("\n\n\n%d\n\n", index);printf("i=%d,j=%d\n", i,j);for (int i = 0; i < 4; i++){for (int j = 0; j < 4; j++){if (i - j == index){sum += arr2[i][j];}}}printf("%d\n", sum);system("pause");sum = 0;}}}system("pause");}

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