C-数字总和

Question

Why won't my program work? It is supposed to add numbers from a formula 1-(1/2)+(1/3)...+(1/999)-(1/1000)=

#include <stdio.h>
#include <math.h>
int main () {
 int i, j;
 float  suma;
 suma = 0.f;
 for (i=0; i<1000; i++) {
    if (i%2==0) {
        suma=suma - 1/i;
    } else {
        suma=suma + 1/i;
    }
 }
 printf("%f", suma);

}

Answer 1

Divide by zero !!

int main () {
 int i;
 float  suma;
 suma = 0.0f; 
 for (i=1; i<1000; i++) {  //fix loop, start from 1
    if (i%2==0) {
        suma=suma - 1.0f/i; // Use 1.0, (1/i will be evaluated as int)
    } else {
        suma=suma + 1.0f/i;
    }
 }
 printf("%f", suma);

}

Answer 2

Try printing 1 / i with i being an int . This will always returns 0, except when i is 1 . This happens because 1 / i is evaluated as an Euclidean division with the remainder being discarded.

The reason this is evaluated like this is because 1 and i are both integer. You need to either the numerator or the denominator to be of floating point type.

One way is to cast i to a float, so your code would look like this: suma = suma - 1 / (float)i . The other way is to make 1 be of floating point type: suma = suma - 1.0 / i or suma = suma - (float)1 / i .

Answer 3

The for loop started from 0. so the first iteration returns divide by zero error. second iteration will return 1/1=1 and will work good, but from third iteration it will return 0, because you are using int. Try starting the for loop from 1 and typecast i to float.

Answer 4

You will get much higher accuracy if you apply some math first. Your series:

1 - 1/2 + 1/3 - 1/4 + ... + 1/999 - 1/1000

can be rewritten as:

(1 - 1/2) + (1/3 - 1/4) + ... + (1/999 - 1/1000)

or as:

1/(1*2) + 1/(3*4) + ... + 1/(999*1000) 

Now, you can write a program to perform calculation. However, you should use double type to improve accuracy and cast integer to double to make sure that your series are added as double numbers:

#include <stdio.h>
#include <math.h>
int main() {
    int i;
    double sum = 0;
    for (i=1; i<1000; i+=2) {
        sum += 1/(i*(i+1.)); // 1. to force cast to double
    }
    printf("%g", sum);
}

Answer 5

herein a simple program

float j = 1.0f;
float suma = 0.0f;
int i = 1;

for (i=1; i <= 1000; i++) {
   suma += j/i;
   j = j * (-1);
}