写出求S=1-1\/2+1\/3-1\/4+...+1\/99-1\/100的程序
\/\/写出求S=1-1\/2+1\/3-1\/4+...+1\/99-1\/100的程序 include "stdio.h"void main(){ int i,n=100;double sum=0;\/\/存储和的变量,初值为0 int sign=1;\/\/控制正负符号的变量 for(i=1;i<=100;i++){ sum=sum+(1.0\/i)*sign;sign=sign*(-1);} printf("S=%lf\\n",sum);...
编写程序求1-1\/2+1\/3-1\/4……+1\/99-1\/100的值
function box(n) { if (n < 1 || n > 100) { return "error" } var result = 0; for (var i = 1; i <= n; i++) { if (i % 2 == 0) { result -= 1\/i; } else { result += 1\/i; } } return result;}console.log(box(100))
用C语言编程:求1-1\/2+1\/3-1\/4+1\/5-1\/6+…+1\/99-1\/100+…的值…
int i,n;\/\/定义循环控制变量和分母值 double sum,k=1;\/\/定义求和变量和转变符号的变量的初值 printf("please input number n:\\n");\/\/提示用户输入想要的分母值 scanf("%d",&n);\/\/需要用户输入的n的值 for(i=1,sum=0;i<=n;i++)\/\/循环从1到n进行求和 { sum+=k\/i;\/\/累加求和 k...
C语言编程计算:1 - 1\/2 + 1\/3 -1\/4 + ... + 1\/99 - 1\/100 +...,直到...
你的第一个问题的程序就是:include <stdio.h> int main(){ int i = 1, sign = 1;double sum = 0, e;do { e = 1.0 \/ i;sum += e * sign;sign = -sign;i++;} while (e > 1e-4);printf("%f\\n", sum);return 0;} ...
编写一个程序求1-1\/2+1\/3-1\/4+…+1\/99-1\/100的值。
include <stdio.h> main(){ int i,k=1;float , sum=0;for(i=1;i<=100;i++){ sum+=k*1.0\/i;k=-k;} printf("sum=%f\\n",sum);}
编写一个Java程序,计算并输出算式1-1\/2+1\/3-1\/4+...+1\/99-1\/100 的...
void main(String[] args) {double a = 1.0;\/\/ 接收最后结果double sum = 0.0; \/\/ 循环 i的值从1-100for (int i = 1; i <= 100; i++) {\/\/ 观察算式,分母是偶数时,分数为负if(i%2 == 0)sum += -(a\/i);elsesum += (a\/i);}System.out.println(sum);}} ...
c语言编程求1-1\/2+1\/3-1\/4+1\/5-1\/6+…….+1\/99-1\/100的值。 可不可以...
{ if(i%2==0) \/\/分母是偶数用减 sum=sum-1.0\/i; \/*这里用 1.0\/i 不能用1\/i,1\/i得到的整型,比如 1\/5会等于0,而这里想要的是等于0.2,所以不对*\/ else \/\/分母是奇数用加 sum=sum+1.0\/i; } printf("结果等于%lf\\n",sum)} ...
1-1\/2+1\/3-1\/4+···+1\/99-1\/100的值是多少
调和数列,没有公式。只能用Excel或编程计算。1-1\/2+1\/3-1\/4+···+1\/99-1\/100≈0.68817
C语言计算1-1\/2+1\/3-1\/4+...+1\/99-1\/100
include<stdio.h> include<math.h> int main(){ int i;float sum=0.0,temp=0.0;for (i = 1; i <= 100; i++){ temp = pow(-1, i + 1);sum = sum + temp * 1 \/ i;} printf("sum is %f\\n", sum);system("pause");return 0;} 运行效果:...
c语言高精度计算 s=1-1\/2+1\/3-1\/4+...+1\/99-1\/100,精确到小数点后100...
} else { overflow = 0; } b[i] %= 10; } CF = overflow;}void Minus(int *b, int *a){ int i; int borrow = 0; for (i=N-1; i>=0; i--) { b[i] -= (a[