用C编写程序，求x=1-1/22+1/333-1/4444+1/55555……+1/999999999

以下是C语言编写的程序，用于计算x=1-1/22+1/333-1/4444+1/55555……+1/999999999：

#include <stdio.h>
#include <math.h>

int fun(int n);

int main() {
    int n, i;
    double sum = 0;
    printf("请输入一个小于9的自然数：");
    scanf("%d", &n);
    for (i = 1; i <= n; i++) {
        sum += pow(-1, i - 1) / fun(i);
    }
    printf("x=%lf\n", sum);
    return 0;
}

int fun(int n) {
    int i, sum = 0;
    for (i = 1; i <= n; i++) {
        sum += n * pow(10, i - 1);
    }
    return sum;
}

相关问题

用C语言编写程序，计算s的近似值，使其误差小于10^-6.(s=1+1/x+1/x^2+1/x^3+1/x^4+……)

#include<stdio.h>
#include<math.h>

int main()
{
double x,sum;
int i;

printf("请输入x的值：");
scanf("%lf",&x);

sum = 1.0; //先将第一项放入sum中
i = 1; //从第二项开始计算

while(1)
{
double term = 1.0/pow(x,i); //计算每一项的值
if(term < 1e-6) //如果当前项的值小于10^-6，则结束循环
{
break;
}
sum += term; //将当前项的值加入到sum中
i++; //计算下一项
}

printf("s的近似值为：%lf\n",sum);

return 0;
}

编写C语言程序，计算s=1+1/x+/x^2+1/x^3……(x>1）的近似值，使其误差小于10的-6次方

#include <stdio.h>
#include <math.h>
int main()
{
double x, s, term;
int n;
printf("Please enter a value of x greater than 1: ");
scanf("%lf", &x);
s = 1.0;
term = 1.0;
n = 1;
while (term >= 1e-6) {
term *= 1.0 / x;
s += term;
n++;
}
printf("s = %.6f\n", s);
printf("The number of terms required is %d\n", n);
return 0;
}