c语言实践 1/1-1/2+1/3-1/4+...

其实这个题目和上面那个是一样的

	/*
	1/1-1/2+1/3-1/4+...1/n;
	*/
	int n = 1;
	double sum = 0;
	double frac = 0;
	int i = 1;

	scanf_s("%d", &n);

	while (i < (n + 1))  //1到n
	{
		if (i % 2 == 0)
		{
			frac = -1.0 / i;
		}
		else
		{
			frac = 1.0 / i;
		}
		
		sum = sum + frac;
		i++;
	}
	printf("sum is %f", sum);

　　

update 2018.10.2

昨天看到一个题目和这个类似。里面不用if判断。

#define N 3
int main(void)
{
    double data;
    int i;
    double sum = 0;
    int mul=1;

    for (i = 0; i < N; i++,mul*=-1)
    {
        data = 1.0 / (i + 1);
        data = data * mul;
        sum = sum + data;
    }

    printf("%f",sum);



    return 1;
}