# include <stdio.h>

int main(void)
{
    float sigh, sum=1, term, i;

    for (i=2; i<=100; ++i)
    {
        term=(-1)*(1/i);
        sum=sum+term;
        ++i;
        sigh=1/i;
        sum=sum+sigh;
    }
    printf("%f\n", sum);

    return 0;
}

结果：0.698073
对么？？？？

# include <stdio.h>

int main(void)
{
   double  sum = 1;
   int  i, sign = 1;

    for (i = 2; i <= 100; i++)
    {
        sign = -1 * sign;
        sum += sign * (1.0 / i);
    }
    printf("%1.10lf\n", sum);

    return 0;
}
我的结果

  1 #include <stdio.h>
  2 int main()
  3 {
  4     int i;
  5     double s=0;
  6     for(i=1;i!=101;++i)
  7     {
  8         s+=1.0/i;
  9         s*=-1;

 10     }

 11     printf("sum = %f\n",s);

 12     return 0;

 13 }

这个貌似是谭浩强老师教材里的一个例子，答案正如2楼，版主的算法也是一样的

#include <stdio.h>
int main(void)
 {
     float sign=1, sum=1, term, i;
 
    for (i=2; i<=100; i++)
     {
          sign=-sign;
         term=(sign)*1/i;         
         sum=sum+term;
     }
     printf("%f\n", sum);
 
    return 0;
 }
 结果是0.688172

最后加了1/101

# include <stdio.h>
int main(void)
{
   
    float sigh, sum=0, term,i;

    for (i=1; i<100; ++i)
    {
        sigh=1/i;
        sum=sum+sigh;
        ++i;
        term=(-1)*(1/i);
        sum=sum+term;
        
        
    }
    printf("%1.12f\n", sum);

    return 0;
}
最后的1/100是负号，加负数最好摆在后面

我的给你看看

#include <stdio.h>

main()
{
    double s=0,m,n;
    int i;
    for(i=1;i<51;i++)
    {
        m=1.0/(2*i-1);
        n=1.0/(2*i);
        s=s+m-n;
    }
    printf("%lf\n",s);
    return 0;
}

#include <stdio.h>

main(){
    int a;
    float b;
    float c=0;
    for(a=1;a<=100;a++){
        if(a%2==0){
            b=-a;
        }
        else{
            b=a;
        }
        c=c+1/b;
    }
    printf("%f",c);
}

结果：0.688172