您好，程序如下
public class BaiDu
{
public static void main(String[] args)
{
int i,n;
double sign=-1.0,t=1.0,sum=0.0;
Scanner read=new Scanner(System.in);
n=read.nextInt();
for(i=1;i<=n;i++)
{
sign=-sign;
t=t*i;
sum+=sign/t;
}
System.out.println("1-1/2!+1/3!-1/4!……+1/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...
public class T {public static 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...
用java编写个1+1\/2+1\/3+1\/4...1\/n
public static void main(String[] args) { System.out.print("输入n："); int n = new Scanner(System.in).nextInt(); double sum = 0; int i = 1; while (i <= n) { sum += (double) 1 \/ i++; } System.out.println(sum);} ...
JAVA编程:已知:s=1-1\/2+1\/3-1\/4+…+1\/(n-1)-1\/n,编写程序求解n=100时的...
public void col(){ double s=1;for(int i=2;i<=100;i++){ if(i%2==0){ s=s-1\/((double)i);}else{ s=s+1\/((double)i);} } System.out.println("result="+s);} 运行结果和你的一致的。
java语言程式设计 输入一个正整数N,输出表达式1+1\/2+1\/3+.+1\/N的值
c语言程式设计。。。解不等式 n < 1+1\/2+1\/3+…+1\/m <n+1，正整数n从键盘输入，m也为正整数，范围待定 include<stdio.h>void main() { int n, i = 1; int min_m =-1, max_m = -1; double sum = 0.0; scanf("%d",&n); while (1){ sum += 1....
jav编程 求1+1\/2!-1\/3!+1\/4!-………+1\/n! n值由键盘接收。
package test;import java.util.Scanner;public class CountTest { \/ 1+1\/2!-1\/3!+1\/4!-1\/5!...+1\/n!return \/ public static int getCount() { int count = 1;int n;Scanner input = new Scanner(System.in);System.out.println("请输入n值！");n = input.nextInt();for (int...
java 求1-1\/2+1\/3-1\/4+1\/5-...
flag=1;for(i=1;i<=n;i++){sum=sum+(float)flag\/i;flag=-flag;} flag是一个符号标记 当为1说明是正数，如果为负说明是负数 (float)flag\/i; 这里相当于 1\/1 -1\/2 1\/3 ...
已知:s=1-1\/2+1\/3-1\/4+…+1\/(n-1)-1\/n,编写程序求解n=100时的S值...
import java.io.IOException;public class JiCheng{ public static void main(String[] args) throws IOException{ double s=0.0000;double x=-1.0;for(int i=1;i<101;i++){ s+=Math.pow(x,i)\/i;} System.out.println("s的值是"+s);} } \/\/本人亲测，完全正确！ 二楼思路完全正确...
用java求1+1\/2+1\/3+……+1\/n
import java.util.Scanner;public class Test40001 { public static void main(String[] args) { int ri, repeat;int i, n;float sum;Scanner in = new Scanner(System.in);repeat = in.nextInt();for (ri = 1; ri <= repeat; ri++) { n = in.nextInt();\/*---*\/ sum = 0F...
编写一个java方法,用来计算并输出1-1\/2+1\/3-1\/4+1\/5-1\/6+...-1\/50...
double sum = 0;int flag = 1;for (int i = 1; i <= 50; i++) { sum = sum + flag * 1.0 \/ i; flag *= -1;}System.out.println(sum);
Java 递归题目,计算1-1\/2+1\/3-1\/4...(-1)^(n+1) *1\/n 要用递归,不要用...
public double foo(int n){ if(n == 1)return 1.0;if(n % 2 == 0)return -1.0\/n + foo(n - 1);else return 1.0\/n + foo(n-1);} 思想就是这么个思想，说要计算分数相加，这块就自己解决吧
