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Finish the function inner_product below. The function should return a[0] * b[0] + a[1] * b[1] + ... + a[n-1] * b[n-1]:

寫inner_product函數。函數會return a[0] * b[0] + a[1] * b[1] + ... + a[n-1] * b[n-1]:

#include <stdio.h>
double inner_product(double a[],double b[],int n)
{
   /*INSERT YOUR CODE HERE*/


   /*END OF YOUR CODE*/
   //The function should return a[0] * b[0] + a[1] * b[1] + ... + a[n-1] * b[n-1].
}

int main(void)
{
   double arrayA[100], arrayB[100];
   int c, n;

   scanf("%d", &n); //Enter number of elements in array
   for ( c = 0 ; c < n ; c++ )     //Enter array A
      scanf("%lf", &arrayA[c]);
   for ( c = 0 ; c < n ; c++ )     //Enter array B
      scanf("%lf", &arrayB[c]);

   printf("%g",inner_product(arrayA,arrayB,n));

   return 0;
}

Example input: 

The first input is the number of elements in arrays (3); The array values of A and B follow; 

第一行是陣列大小,第二行A元素,第三行B元素

3
4 7 8
1 2 3

Example output

42

Remember: You may correct the cases, but your code always be revised!

[Exercise] Coding (C) - asked in Chapter 9: Functions by (5.2k points)
ID: 37277 - Available when: 2017-12-14 18:00 - Due to: Unlimited

edited by | 877 views

38 Answers

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answered by (-32 points)
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answered by (-281 points)
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Case 1: Correct output
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answered by (-108 points)
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