#include<stdio.h>
int main()
{
float sum=0.0;
int a=1;
for(int i=1;i<=100;i++)
{
sum=sum+a*(1.0/i);
a=-a;
}
printf("sum=%f\n",sum);
return 0;
}
原创
2022-06-29 20:05:11
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思路是flag改变符号#define _CRT_SECURE_NO_WARNINGS 1#include<stdio.h>int main(){ double sum = 0.0; int i = 0; int flag = 1; for (i = 1; i <= 100; i++) { sum += flag*1.0 / i; flag = -flag;
原创
2022-06-09 10:18:33
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满意答案1: Dim i As Integer Dim s As Double Dim s0 As Double For i = 1 To 100 s0 = 1 / i If i Mod 2 = 0 Then s0 = s0 * (-1) s = s + s0 Next Print s
2: Dim i As Integer Dim s As Double For i = 1 To 100 Ste
转载
2023-07-02 13:21:05
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#include<stdio.h>
#include<math.h>
int main()
{
double i=1.0,sum=0.0;
for(i=1.0;i<=100.0;i++)
{
sum=sum+(1/i)*pow(-1,i+1);
}
printf("%f\n",sum);
return 0;
}
原创
2015-10-07 18:13:04
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1、正负号如何切换
2、用浮点型去计算
原创
2023-02-27 12:37:51
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/******************************************* 求1-1/2+1/3-1/4+... -1/100的值*******************************************/#include <stdio.h>int main(void){ int i = 1; float sum = 0; while(
原创
2022-08-17 14:29:52
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#include<stdio.h>int main(){ int i = 1; int flag = 1; double sum = 0; for (i = 1; i <= 100; i++) { sum += flag * (1.0 / i); flag = -flag; } printf("总共的和为%lf", sum); return 0;
原创
2022-12-31 21:26:21
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#include<stdio.h>int main(){ int i = 1; int flag = 1; double sum = 0; for (i = 1; i <= 100; i++) { sum += flag * (1.0 / i); flag = -flag; ...
原创
2023-01-21 15:09:50
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package ch22;public class CalculateDemo { public static void main(String[] args) { int n=100;// 循环上限 double sum=0; for(double i=1;i<=n;i++){ if(i%2==0) sum=sum-(1/i);//如果i为偶数，则减 ...
原创
2021-08-21 20:17:44
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package ch22;public class CalculateDemo { public static void main(String[] args) { int n=100;// 循环上限 double sum=0; for(double i=1;i<=n;i++){ if(i%2==0) sum=sum-(1/i);//如果i为偶数，则减 ...
原创
2022-03-04 15:07:10
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#每日美图分享##include<stdio.h>int main(){ int num = 1; int flag = 1; double sum = 0.0; for (num = 1; num <= 100; num++) { sum += flag * (1.0 / num); flag = -flag; } printf("%lf\n", su
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精选
2022-10-12 22:34:08
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#include<stdio.h>
int main()
{
int sign = 1;
double deno = 2.0, sum = 1.0, term;//deno分母，sum多项式，term分式
while (deno <= 100)
{
sign = -sign;
原创
2022-09-29 23:19:47
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C语言：计算1/1-1/2+1/3-1/4+1/5 …… + 1/99 - 1/100 的值
原创
2015-10-07 17:05:03
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1评论
注意:计算时1要用double类型即1.0。奇数偶数分开计算，再合并。#include<stdio.h>
int main()
{
int i;
double sum=0,sum1=0,sum2=0;
for(i = 1;i <= 99;i+=2)
{
sum1=sum1+1.0/i;
原创
2015-10-11 10:34:53
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#include<stdio.h>#include<math.h>intmain(){doublen;doublesum=0.0;for(n=1;n<=100;n++){sum+=(1/n)*pow(-1,n-1);}printf("%sum=lf",sum);return0；}这里可以有更好的选择，降低运行时间，也可以不用到pow函数。#include<std
原创
2020-03-18 18:38:33
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#include<stdio.h>
int main()
{
int i =0;
double sum =0.0;
int flag = 1;
for(i=1; i<=100; i++) //i不能等于0 i是除数
{
sum += flag*1.0/i; //如果把1.0写成1 结果就等于1 1除2 商0 只有第一个
原创
2021-01-19 22:21:39
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#include <stdio.h>
#include <stdlib.h>
int main()
{
int sign = 1;
double deno = 2.0;
double sum = 1.0;
原创
2015-12-05 14:59:05
1597阅读
先来看一种错误写法：此时有两处问题：1.sum返回值应该是个小数，所以用int不对；2.这个循环是纯加法，不满足题目要求；改进如下：第一个问题改进后，同实参a来确定sum正负：okok搞定了！
原创
2021-11-28 19:35:06
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c语言：求多项式1-1/2+1/3-1/4+...+1/99-1/100的值，3种循环实现
原创
2015-11-29 01:06:36
10000+阅读
#include<stdio.h>int main(){ int i = 0; double num = 0.0; int flag = 1; for (i = 1; i <= 100; i++) { num += (1.0 / i)*flag; flag = -flag; &
原创
2015-12-06 16:44:13
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