数列求和计算

1. $$f\left(n\right)=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots +\frac{1}{\mathrm{n}}$$

#include <stdio.h>
int main() {int i;int n;double sum = 0.0;printf("请输入n的值：");scanf("%d",&n);getchar();for (i = 1; i <= n; i++){sum += 1.0/i;}printf("f(%d)=%f",n,sum);getchar();return 0;
}

2. $$f\left(n\right)=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\dots -\frac{1}{\mathrm{n}}$$

#include <stdio.h>
int main() {int i;int n;int sign = 1;double sum = 0.0;printf("请输入n的值：");scanf("%d",&n);getchar();for (i = 1; i <= n; i++){sum += sign*1.0/i;sign = -sign;}printf("f(%d)=%f",n,sum);getchar();return 0;
}

定义一个整数变量来满足程序中的+和-的变换。
当然也可以直接定义一个double变量来实现

#include <stdio.h>
int main() {int i;int n;double sign = 1.0;double sum = 0.0;printf("请输入n的值：");scanf("%d",&n);getchar();for (i = 1; i <= n; i++){sum += sign/i;sign = -sign;}printf("f(%d)=%f",n,sum);getchar();return 0;
}