C-1 / n数的总和

Question

Hello could you take a quick glance at my code and point out the mistake(s). I'm trying to calculate sum of n numbers going like this: 1- 1/2 + 1/3 - 1/4 ... etc...

With the following code, I get 1.00000 each time, but it should be between 0 and 1, for example for 3 it should be 1 - 1/2 + 1/3 = 0,83333.

#include <stdio.h>
int main () {

 int n, prefix;
 float sum;
 scanf("%d", &n);
 do {
   if (n%2==0) {
    prefix=-1;
   } else {
    prefix=1;
   }
   sum+=  prefix/n;
   n = n - 1;

 } while (n > 0);
 printf("%f", sum);

}

Answer 1

Three errors that I can see:

1. typo ( premix vs prefix )
2. use of integer type
3. (thanks to @Light): initialization of sum

Try the following instead:

#include <stdio.h>
int main () {

 int n, n_initial;
 double sum=0.0, prefix = 1.0;
 printf("enter the value for n:\n");
 scanf("%d", &n);

 if(n<1) {
   printf("n must be > 0!\n");
   return 1;
 }
 n_initial = n;
 if (n%2==0) prefix = -1.0; else prefix= 1.0;
 do {
   sum+= prefix/(double)n;
   prefix *= -1.0;
   n--;

 } while (n > 0);
 printf("The sum of the series over %d terms is: %lf\n", n_initial, sum);
 return 0;
}

Note - I keep n as an integer, and cast it explicitly before the divition. It might be better just to make it a float / double - remember to change the format specification for the scanf accordingly. I only do the modulo operation once (after that, the sign of prefix just keeps changing). Also - it's always a good idea to add a prompt for the input of a number ("why isn't it doing anything?!"), to annotate the result (rather than print "just a number"), and to end the output with a newline (so the prompt doesn't obscure the output of the program).

Finally - you might want to check that the user doesn't enter a negative number, which would make your code give a bad result.

As an afterthought - you could test for large values of n , and just return log(2.0) . But that would be cheating... and this series does converge awfully slowly (it oscillates quite badly - the 3rd digit is still changing when n=1000). Consequently, rounding errors risk really compounding. This is why you need to be using the double type; but I would suggest it's instructive to look at other ways to compute log(2.0) - for example, using one of the other series given at http://www.math.com/tables/expansion/log.htm . You could actually implement all of them, and compare their accuracy after n terms (by printing out the error: sum - log(2.0) .)

Answer 2

在上面的答案中加上未初始化的总和 （将其初始化为零 ）。并且我认为您的代码在输入为0时不起作用（它给出除以零的误差） 。因此，最好在while或for循环中使用或提出一个在做一会儿。

Answer 3

you are doing prefix/n when both are integers. so you are getting the answer as integer (which always is 0)

use

(double)prefix/n

or

prefix/(double)n

or even

(double)prefix/(double)n

here is the full code:

#include <stdio.h>
int main () {

 int n, prefix;
 float sum = 0;
 scanf("%d", &n);
 do {
   if (n%2==0) {
    prefix=-1;
   } else {
    prefix=1;
   }
   sum+= (double)prefix/(double)n;
   n = n - 1;

 } while (n > 0);
 printf("%f", sum);

}

i ran it on compileonline.com with the input of 10 and got:

Answer 4

Both prefix and n are integers, so that means you will have integer division which truncates the result (ie simply cuts of the decimals). If you want a floating point result, one or both of the variables also have to be a floating point variable.