#include"stdio.h"
main()
{ float a=2.0,b=1.0,sum=0,t,i;
for(t=0,i=1;i<=3;i++)
{ t=a/b;
b=a;
a=a+b;
sum=sum+t;
}
printf("%f",sum);
}

你要什么结果？

浮点数好不好

printf("%.3f",sum);

~~~~~~~~~~~~~~~~~~~~~~~~~~

因为都能整除啊

第一次循环
t=a/b    t=2.0/1.0=2.0   t=2.0
b=a   a==2.0                b=2.0
a=a+b  a=2.0+2.0           a=4.0
sum=sum+t    sum=0+2.0     sum=2.0

第二次循环
t=a/b    t=4.0/2.0     t=2.0
b=a    a==4.0           b=4.0
a=a+b   a=4.0+4.0       a=8.0
sum=sum+t    sum=2.0+2.0    sum=4.0

第三次循环
t=a/b    t=8.0/4.0     t=2.0
b=a     a==8.0         b=8.0
a=a+b    a=8.0+8.0     a=16.0
sum=sum+t    sum=4.0+2.0   sum=6.0

root@~ #cat 3.c
#include <stdio.h>

int main (void) {

        float a=2.0,b=1.0,sum=0,t,i;

        for(i=1;i<=3;i++) {
                t=a/b;
                printf ("a=%f b=%f t=%f",a,b,t);
                b=a;
                a=a+b;
                sum=sum+t;
                printf (" sum=%f\n",sum);
        }

        printf ("Result is %f\n",sum);

        return 0;

}

root@~ #./3
a=2.000000 b=1.000000 t=2.000000 sum=2.000000
a=4.000000 b=2.000000 t=2.000000 sum=4.000000
a=8.000000 b=4.000000 t=2.000000 sum=6.000000
Result is 6.000000
root@~ #