9.编写一个程要求用户输入下限整数和一个上限整数,然后。依次计算从下限到上限每一个整数的平方的和,最后显示结果。程序将不断提示用户输入下限整数和上限整数并显示出答案,直到用户输入上限等于或小于下限整数为止,程序运行的结果示例应该如下所示
Enter lower and uppe integer linits:5 9
The sums of the squares from 25 to 81 is 255
Enter lower and uppe integer linits:5 5
Done
//6-16-9.c
#include <stdio.h>
int main (void)
{
long a,b,sum = 0; //b上限,a下限
printf ("Enter lower and uppe integer linits:");
for (scanf ("%ld %ld",&a,&b);b > a;)
{
printf ("The sums of the squares from %ld to %ld is ",a * a,b * b);
while (b >= a)
{
sum = sum + a * a;
a++;
}
printf("%ld\n",sum);
printf ("Enter lower and uppe integer linits:");
scanf ("%ld %ld",&a,&b);
}
printf ("Done\n");
return 0 ;
}
10.编写一个程序把8个整数读入一个数组中,然后相反的顺序打印它们
//6-16-10.c
#include <stdio.h>
#define GS 6
int main (void)
{
int a;
int b[GS] ;
for (a = 0 ;a < GS;a++)
scanf ("%d",&b[a]);
for (a = 0 ;a < GS;a++)
printf("%5d",b[a]);
printf ("\n");
for (a = a-1; a >= 0;a--)
printf("%5d",b[a]);
printf ("\n");
return 0 ;
}
11.考虑这两个无限序列:
1.0+1.0/2.0+1.0/3.0+1.0/4.0+…
1.0-1.0/2.0-1.0/3.0-1.0/4.0-…
编写一个程序来计算两个序列不断变化的总和,直到达到某个次数,让用户交互地输入这个次数看看到20次,100次和500次后的总各,是否每个序列都年上去要收敛于某个值?提示奇个-1相乘的值为-1而偶个数相乘的值的1。
//6-16-11.c
#include <stdio.h>
int main (void)
{
int a,c = 1;
float b,e,sum = 0, sum1 = 0;
while (scanf ("%d",&a) == 1)
{
for (b = 1.0;(int)b < a;b++ )
{
sum += (1 / b);
printf ("%.3f\t",sum);
}
printf ("\n");
printf ("第一个序列的和:%.5f\n",sum);
sum = 0;
for (b = 1.0;(int)b < a; b++)
{
sum += (1 / b) * c;
c *= -1;//这点我没有想到
printf ("%.3f\t",sum);
}
printf ("\n");
printf ("第二个序列的和:%.5f\n",sum);
sum = 0;
}
return 0 ;
}
12.编写一个程序,创建一个8个元素的int数组,并且把元素分别设置为2的前8次幂,然后打印出它们的值,使用for循环来设置值,为了变化,使用do while循环来显示这些值。
#include <stdio.h>
#define MZ 8
int main (void)
{
int a[MZ];
int b = 1,c,d = 1;
for (c = 0;c < MZ;c++)
{
a[c] = d;
d = d * 2;
do
{
printf("%d\n",a[c]);
b ++;
}
while (b < c);
}
return 0 ;