已知一维数组a[9]={1,2,3,4,5,6,7,8,9},求该数组中下标为奇数元素之和...答:下标为奇数元素,即下标是 i=1,3,5,7, 的元素 a[i] 之和。用循环语句,步长为2:for (i=1;i<9;i=i+2) 。。。求和 用: sum=sum+a[i];累加器 初值给0。程序如下:include<stdio.h> int main(){ int a[9]={1,2,3,4,5,6,7,8,9};int i,sum;sum=0;for (i=1;i<...
求一维数组A中值为奇数的元素之和答:int sum( int b[ ],int n ){ int i,s = 0;for ( i=0; i<n; i++)if (b[i] % 2 == 1)s = s + b[i]return (s);} main(){ int a[12]={10,4,2,7,3,12,5,34,5,9,21,19},n;clrscr();n = sum(a,12);//这里应该改为n=sum(a,12);printf("The result...