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1.题目要求
一个5*5的矩阵,求出矩阵两条对角线上的各元素之和
1.这道题的重点在于要清楚知道主对角线和副对角线上的各元素的下标特征,把这个个问题弄清楚就不难了
2.不理解的可以看一下例图:
2.解题思路
由上图可知:
i是行,j是列,第一条红色主对角线上的元素下标很容易就能看出其规律,那就可以将a[i][j]放在循环里,重复进行五次++就可以得到他的元素.
第二条蓝色对角线上的元素,可以定义一个变量n,赋上列的值也就是4,行号仍用i来定义.
注意:两条对角线上有一个重复的值,可以将它减去一次.
代码如下（示例）：
#include<stdio.h> int main() { int sum = 0;//存放累加的值 int arr[][5] =
{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25}; for
(int i=0,j=0, n=4;i<5&&j<5&&n>=0;i++,j++,n--)//用n来遍历副对角线的值 { sum +=
arr[i][j] + arr[i][n]; } sum -= sum - arr[2][2];//减去一次重复的值
printf("%d\n", sum); return 0; }
<>总结
如有不足,还请指出,期待和各位一起进步
