#include <stdio.h>
main ()
{
float a,b;
float sum=0,term;
do
{
a=2;
b=1;
term=a/b;
sum=sum+term;
a=a+b;
b=a-b;

}
while(a<5);
printf("the number is %.15f",sum);
}
题目的意思是  求2/1+3/2+5/3+8/5+……

main ()
{
float a,b;
float sum=0,term;
do
{
a=2;
b=1;
term=a/b;
sum=sum+term;
a=a+b;
b=a-b;

}
while(a<5);
printf("the number is %.15f",sum);
}

#include <stdio.h>
main ()
{
float a,b;
float sum=0,term;
do
{
a=2;
b=1;
term=a/b;
sum=sum+term;
a=a+b;
b=a-b;

}
while(a<5); //解释下a<5是什么意思？？只算到2/1+3/2+5/3？？？？？？
printf("the number is %.15f",sum);
}

#include <stdio.h>

int
main (void)
{
      float a=2,b=1;
      float sum=0,term;
      do
      {
            term=a/b;
            sum=sum+term;
            a=a+b;
            b=a-b;
      }
      while(b<=5);
      printf("the number is %.15f",sum);
      getch();
      return 0;
}

计算到2/1+3/2+5/3+8/5

可不可以考虑一下数组啊？

我想用数组解决，但我编的程序是错的，
#include <stdio.h>
#define N 10000
int
main (void)
{
      int f[N],g[N],term[N],h[N];
      float sum=0,n,a;
      f[0]=2;g[0]=1;h[0]=0;
      for(n=0;n<=10000;n++)
      {
      f[n+1]=f[n]+g[n];
      g[n+1]=f[n]-g[n];
      term[n]=f[n]/g[n];
      h[n]=h[n-1]+term[n];
      }
      scanf ("%d",&a);
      printf("%d",h[a]);
}
师哥们帮帮忙改一下！

解决了 哈哈！