计算表达式:s=1-2/3+4/5-6/7+8/9-10/11+……-98/99+100/101并输出,要求结果保留5位小数
代码如下:
#include <iostream>
#include<iomanip>
using namespace std;
int main()
{
cout<<"s="<<1;
double sum=1.0;
int count = 1;
for(int i=2;i<=100;i=i+2)
{
if(count%2==0)
{
cout<<"+";
sum+=1.0*i/(i+1);
}else{
cout<<"-";
sum-=1.0*i/(i+1);
}
cout<<i<<"/"<<i+1;
count++;
}
cout<<"=";
cout<<fixed<<setprecision(5)<<sum<<endl; //保留5位小数
return 0;
}
有用请采纳!
#include <stdio.h>
int main()
{
int pos = -1;
float result=1.0;
//int n;
//scanf("%d",&n);
printf("1");
for(int i=2;i<=100;i+=2)
{
result = result + (float)(pos * i )/ (i +1);
printf(" + (%d/%d) " , pos*i,i+1);
pos = -pos ;
}
printf("\r\nresult = %.5f\r\n",result);
}