【C/C++程式碼練習11】1-1/2+1/3-1/4+...+1/n的兩種計算方法
阿新 • • 發佈:2019-01-01
方法一:
#include <stdio.h> #include <math.h> int main(void) { int n; scanf("%d", &n); int i; double sum = 0; double sign; for (i=1; i<=n; i++) { sign = pow(-1, i+1); sum = sum + 1.0/(sign*i); } printf("f(%d) = %f\n", n, sum); return 0; }
方法二:
#include <stdio.h>
int main(void)
{
int n;
scanf("%d", &n);
int i;
double sum = 0;
double sign = 1.0;
for (i=1; i<=n; i++)
{
sum = sum + sign/i;
sign = -sign;
}
printf("f(%d) = %f\n", n, sum);
return 0;
}