c++程序设计设s=1+1/2+1/3+....+1/n,求与八最接近的s的值与其对应的n值。最好用到do。。。while语

声明一个int型变量n作为分母兼作项数记数变量，声明一个double型变量s记录前n项之和；利用do~while语句，n从1开始到当前项的浮点和小于8时继续循环，否则退出。退出循环后，比较s-8和8-(s-1.0/n)，取小者(相等时取前一项)的最后一项的n便是题解。代码如下：
#include "stdio.h"int main(int argc,char *argv[]){int n;double s;s=n=0;do{s+=1.0/++n;}while(s 8-s+1.0/n)s-=1.0/n--;printf("s = %fn = %d
",s,n);return 0;}运行结果如下：

#include
int main()
{ int i=0;
double s=0;
while(s<8)
s+=1.0/++i;
if(s-8<8-(s-1.0/(i-1)))
printf("s=%fi=%d
",s,i);
else printf("s=%fi=%d
",s-1.0/(i-1),i-1);
return 0;
}

#include<iostream>
#include<windows.h>
using std::cout ;
using std::endl ;
int main()
{
int n=1 ;
double s=0,s2 ; //s是当前表达式的和，s2是上一次计算时的和。
do{
s2=s ;
s+=(double)(1.0/n) ;
n++ ;
}while(s<8) ; //退出循环的时候，s大于8，s2小于等于8
if(8-s2>s-8) //算绝对值。。。小的输出
cout<<"s="<<s<<" n="<<n-1<<endl ;
else cout<<"s="<<s2<<" n="<<n-2<<endl ; //输出
return 0 ; //return 0；
}

#include<iostream>
using namespace std;
void main()
{
int n=1;
double s=0;
do
{
s=s+(1.0/n);
n++;
}
while(s<8.0);
cout <<"s="<<s<<endl;
cout <<"n="<<n<<endl;
}

#include<cstdio>
int main()
{
double s1=0,s2;
int t=0;
do {
t++;
s2=s1;
s1+=1.0/t;
}while(s1<8);
if(s1+s2>16) printf("%lf %d\n",s2,t-1);
else printf("%lf %d\n",s1,t);
}
补：
#include<cstdio>
#include<cmath>
int main()
{
int n;
while(~scanf("%d",&n)){
double s1=0,s2;
int t=0;
do {
t++;
s2=s1;
s1+=1.0/t;
}while(s1<8&&t<n);
if(fabs(s1-8)>fabs(8-s2)) printf("%lf %d\n",s2,t-1);
else printf("%lf %d\n",s1,t);
}
}

#include<iostream>
using namespace std;
int main()
{
double sum=0;
double i=1;
do
{
sum+=(1.0/i);
i+=1;
}while(sum<8.0);
i-=1;
double temp = sum-1.0/i;
double a = 8-temp;
double b = sum-8;
if (a<b)
{
sum = temp;
i-=1;
}
cout<<i<<"\t"<<sum<<endl;
}

#include<iostream>
void main()
{
int n=1;
double limit;
double sum=0;
cin>>limit;//输入limit
while(1)
{
sum=sum+(double)(1/n);//求和
if (sum>limit)//满足条件
{
cout<<n;//输出n
return;//跳出循环
}
n++;
}
return;
}

求助!!c++程序设计设s=1+1\/2+1\/3+...+1\/n,求与八最接近的s的值与其对...
do{ s2=s ;s+=(double)(1.0\/n) ;n++ ;}while(s<8) ; \/\/退出循环的时候，s大于8，s2小于等于8 if(8-s2>s-8) \/\/算绝对值。。。小的输出 cout<<"s="<<s<<" n="<<n-1<<endl ;else cout<<"s="<<s2<<" n="<<n-2<<endl ; \/\/输出 return 0 ; \/\/retur...

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C语言编程,s=1+1\/2-1\/3+1\/4-1\/5...+1\/n,
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c++程序设计设s=1+1\/2+1\/3+...+1\/n,求与八最接近的s的值与其对应的n值...
include "stdio.h"int main(int argc,char *argv[]){int n;double s;s=n=0;do{s+=1.0\/++n;}while(s<8);if(s-8 > 8-s+1.0\/n)s-=1.0\/n--;printf("s = %f\\tn = %d\\n",s,n);return 0;}运行结果如下：

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C++编写程序,计算s=1+(1+2)+(1+2+3)+…+(1+2+3+…+n)的值
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