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试题描述
求一个3×3的整型二维数组对角线元素之和。请使用数组实现。
输入
输入包含3行，每行均有3个整数，为二维数组按行列顺序输入的元素。相邻两项之间用一个空格隔开。
输出
输出对角线元素之和。
输入示例
1 2 -3
4 5 6
-7 8 9
输出示例
15
数据范围
输入输出均为int范围的整数
根据题意，输入为每行三个数，为三行，显然需要定义3x3二位数组
#include <stdio.h> #include <stdlib.h> int main() { int a[3][3],i,m,j,sum=0;
for(i=0;i<3;i++){ scanf("%d %d %d",&a[i][0],&a[i][1],&a[i][2]); }
题意为输出主对角线和的值，观察发现主对角线的特点，横纵坐标一致，
即[0][0] ，[1][1] ，[2][2] ，所以条件即为横纵坐标相等。
for(m=0;m<3;m++){ for(j=0;j<3;j++){ if(m==j){ sum=sum+a[m][j]; } } }
printf("%d",sum); return 0; }
根据条件挨个检索，符合加和即可
综上最终代码即为
#include <stdio.h> #include <stdlib.h> int main() { int a[3][3],i,m,j,sum=0;
for(i=0;i<3;i++){ scanf("%d %d %d",&a[i][0],&a[i][1],&a[i][2]); }
for(m=0;m<3;m++){ for(j=0;j<3;j++){ if(m==j){ sum=sum+a[m][j]; } } }
printf("%d",sum); return 0; }
（代码，解法不唯一，仅供参考)
