第五章习题源程序.doc

第五章习题参考源程序
5.1输入 入个正整数M和N,求其中最大公约数和最小公倍数
#include <math.h>
#include <stdio.h>
main()
{ unsigned int m,n,k,maxgy,mingb;
int i;
printf("Please input the two number of m,n\n");
scanf("%u,%u",&m,&n);
if (m>n)
k=n;
else
k=m;
for (i=k;i>1;i--)
{if ((m%i==0)&&(n%i==0))
break;
}
printf("%u\n",i);
if (i>1)
{printf("These two number have maxgy=%u\n",i);
mingb=i*(m/i)*(n/i);
printf("The two number have mingb=%u\n",mingb);
}
else
{printf("These two number have maxgy=%d\n",i);
printf("These two number have mingb=%u\n",m*n);}
}

5.2输入一行字符,分别统计出其中英文字母、空格、数字和其他字符的个数
#include <math.h>
#include <stdio.h>
main()
{ char c;
int num_count=0,char_count=0,space_count=0,other_count=0;
while ((c=getchar())!='\n')
{
if ((c>='a' && c<='z') || (c>='A' && c<='Z'))
char_count++;
else if (c>='0' && c<='9')
num_count++;
else if (c==32)
space_count++;
else
other_count++;
}
printf("The number of character is:%d\n",char_count);
printf("The number of charcter is :%d\n",num_count);
printf("the number of space is :%d\n",space_count);
printf("The number of other symbol is :%d\n",other_count);
}
5.3 求Sn=a+aa+aaa+….aaaaaa之值，其中a是一个数字。例如：2+22+222+2222+22222（此时n=5）,n由键盘输入。
#include <math.h>
#include <stdio.h>
main()
{ short int a,b,n,i;
long int sum=0;
scanf("%1d,%2d",&a,&n);
b=a;
for (i=1;i<=n;i++)
{ sum+=a;
a=a*10+b;
}
printf("The result of this equation is :%d\n",sum);

}
5.4 求
#include <stdio.h>
main()
{
int i,j;
int n;
long int m, sum=0;
printf("Please input the number of N\n");
scanf("%d",&n);
for (i=1;i<=n;i++)
{m=1;
for (j=1;j<=i;j++)
m*=j;
sum+=m;
}
printf("The number of N is :%d\n",n);
printf("The sum of N! is :%ld\n",sum);
}
5.5 打印出所有的“水仙花数”，所谓：“水仙花数”是指一个3位数，其各位数字立方和等腰三角形于该数本身。例如153是一水仙花数，因为 153=13+53+33
#include <math.h>
#include <stdio.h>
main()
{ short int c1,c2,c3;
int n;
for (n=100;n<=999;n++)
{c3=n/100;
c2=(n-c3*100)/10;
c1=(n-c3*100-c2*10)/1;
if (n==c3*c3*c3+c2*c2*c2+c1*c1*c1)
printf("The number is a shuixian flower :%d\n",n); }
}

5.6 一个数如果恰好等腰三角形于它的因子之各，这个数就称为“完数”。例如，6的因子是1，2，3，而6=1+2+3，因些6是：“完数”。编程序找出1000之内的所有完数，并按下面格式输出其因子。
6 is factors are 1,2,3

#include <math.h>
#include <stdio.h>
main()
{
int n,i,sum;
for (n=2;n<1000;n++)
{ sum=0;
for (i=1;i<n;i++)
if (n%i==0)
sum+=i;
if (n==sum)
{printf(" %d is factors are:",n);
for (i=1;i<n;i++)
if (n%i==0)
printf(" %d,",i);
printf("\n");
}
}
}

5.7 有一个分数序列 求出这个数列的前20项之和。
#include <math.h>
#include <stdio.h>
main()
{ int n;
float a1=1;
float a2=2;
float b1=2;
float b2=3;
float sum=0.0;
sum+=b1/a1;
printf("1 sum of %f\n",sum);
sum+=b2/a2;
printf("2 sum of %f\n",sum);
for (n=3;n<12;n++)
{ a1=a1+a2;
b1=b1+b2;
sum+=b1/a1;
printf(" %d sum of %f\n",n,sum);
a2=a2+a1;
b2=b2+b1;
sum+=b2/a2;
printf(" %d sum of %f\n",n,sum);
}
}
5.8 猴子吃桃子问题。猴子第一天摘下若有干个桃子，当即吃了一半，还不过瘾，又多吃了一个。第二天早上又将剩下的桃子吃掉了一半，又多吃了一个。以后每天早上都吃了前一天剩下一的一半零一个。到第10天早上想吃时，就只剩下一个桃子了。求第一个共摘多少桃子。
#include <math.h>
#include <stdio.h>
main()
{ int sum=1;
int n;
printf("The number of peachsis \n");
for(n=1;n<10;n++)
{
sum=(sum+1)*2;
printf("The %d day peach is:",10-n);
printf("%d\n",sum);
}
}
5.8 用迭代法求。求平方根的迭代公式为，要求前后两次求出的x的差的绝对值小于10-5。
#include <math.h>
#include <stdio.h>
main()
{
float a;
double x1,x2;
printf("Please input the number of a\n");
scanf("%f",&a);
if (fabs(a)<=1e-6)
printf("The number you input must a positive number\n");
else
{ x1=1.0; /*注意：在进行实数运算时，1等数要写成1.0*/
printf("x1 %f\n",x1);
x2=1.0/2*(x1+a/x1);
printf("x2 %f\n",x2);
while(fabs(x2-x1)>1e-5)
{
x1=1.0/2.0*(x2+a/x2);
printf("x1 %f\n",x1);
x2=1.0/2.0*(x1+a/x1);
printf("x2 %f\n",x2);}

}
printf("the value of sqrt(a) is :%10.4f\n",x1);
}

5.10打印出以下图案

*
* * *
* * * * *
* * * * * * *
* * * * *
* * *
*
#include <math.h>
#include <stdio.h>
main()
{
int n;
int i,j;
printf("please input the line number:");
scanf("%d",&n);
for (i=1;i<=n;i++)
if (i<=(n+1)/2)
{for (j=1;j<=(n+1)/2-i;j++)
printf(" ");
for (j=1;j<=2*i-1;j++)
printf("*");
printf("\n"); }
else

{ for (j=1;j<=i-(n+1)/2;j++)
printf(" ");
for (j=1;j<=n-(i-(n+1)/2)*2;j++)
printf("*");
printf("\n");
}

}
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