#include<stdio.h>
int main(void)
{
    double a;
    double b;
    double c=1.23456789012345678901234567980123456789;

    float  d;
    float  e;
    float  f=1.23456789012345678901234567890123456789;
   
    printf("输入a=");  scanf("%f",&a);
    printf("输入b=");  scanf("%lf",&b);

    printf("输入d=");  scanf("%f",&d);
    printf("输入e=");  scanf("%lf",&e);

    printf("a=%.33f\n",a);
    printf("b=%.33f\n",b);
    printf("c=%.33f\n",c);
    printf("d=%.33f\n",d);
    printf("e=%.33f\n",e);
    printf("f=%.33f\n",f);

    printf("a=%f\n",a);
    printf("b=%f\n",b);
    printf("c=%f\n",c);
    printf("d=%f\n",d);
    printf("e=%f\n",e);
    printf("f=%f\n",f);
}

运行结果：
输入a=1.23456789012345678901234567890123456789
输入b=1.23456789021245678901234567890123456789
输入d=1.23456789012345678901234567890123456789
输入e=1.23456789012345678901234567890123456789
a=-92559604304082489000000000000000000000000000000000000000000000.00000000000000
0000000000000000000
b=1.234567890212456800000000000000000
c=1.234567890123456700000000000000000
d=1.904320955276489300000000000000000
e=70.175743103027344000000000000000000
f=1.234567880630493200000000000000000
a=-92559604304082489000000000000000000000000000000000000000000000.000000
b=1.234568
c=1.234568
d=1.904321
e=70.175743
f=1.234568
Press any key to continue

求解释1.输出的结果a
      2.各个结果后面的0的个数
      3.double型的输入不能用scanf("%f",&x)?   
      4.第一组b和c的结果为什么不一样而第二组的b和c结果一样
      5.d、e、f的结果
      6.等等

1与3.double型的输入应该用scanf("%lf",&x)
2与4.double的精度应该就15  16 位

为什么后面那么多0，结果的有效数字不是17位么？