9.編寫一個程要求用戶輸入下限整數和一個上限整數,然後。依次計算從下限到上限每一個整數的平方的和,最後顯示結果。程序將不斷提示用戶輸入下限整數和上限整數並顯示出答案,直到用戶輸入上限等於或小於下限整數爲止,程序運行的結果示例應該如下所示
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.編寫一個程序把8個整數讀入一個數組中,然後相反的順序打印它們
//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.考慮這兩個無限序列:
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-…
編寫一個程序來計算兩個序列不斷變化的總和,直到達到某個次數,讓用戶交互地輸入這個次數看看到20次,100次和500次後的總各,是否每個序列都年上去要收斂於某個值?提示奇個-1相乘的值爲-1而偶個數相乘的值的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 ("第一個序列的和:%.5f\n",sum);
sum = 0;
for (b = 1.0;(int)b < a; b++)
{
sum += (1 / b) * c;
c *= -1;//這點我沒有想到
printf ("%.3f\t",sum);
}
printf ("\n");
printf ("第二個序列的和:%.5f\n",sum);
sum = 0;
}
return 0 ;
}
12.編寫一個程序,創建一個8個元素的int數組,並且把元素分別設置爲2的前8次冪,然後打印出它們的值,使用for循環來設置值,爲了變化,使用do while循環來顯示這些值。
#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 ;