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 #include main() unsigned int m,n,k,maxgy,mingb; int i; printf(Please input the two number of m,nn); scanf(%u,%u,&m,&n); if (mn) k=n; else k=m; for (i=k;i1;i-) if (m%i=0)&(n%i=0) break; printf(%un,i); if (i1)printf(These two number have maxgy=%un,i); mingb=i*(m/i)*(n/i); printf(The two number have mingb=%un,mingb); else printf(These two number have maxgy=%dn,i); printf(These two number have mingb=%un,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 #include main() char c; int num_count=0,char_count=0,space_count=0,other_count=0; while (c=getchar()!=n) if (c=a & c=A & c=0 & c=9) num_count+; else if (c=32) space_count+; else other_count+; printf(The number of character is:%dn,char_count); printf(The number of charcter is :%dn,num_count); printf(the number of space is :%dn,space_count); printf(The number of other symbol is :%dn,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 #include 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 :%dn,sum); 5.4 求#include main() int i,j;int n;long int m, sum=0;printf(Please input the number of Nn);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 :%dn,n); printf(The sum of N! is :%ldn,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 #include 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 :%dn,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 #include main() int n,i,sum; for (n=2;n1000;n+) sum=0; for (i=1;in;i+) if (n%i=0) sum+=i; if (n=sum) printf( %d is factors are:,n); for (i=1;in;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 #include 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 %fn,sum); sum+=b2/a2; printf(2 sum of %fn,sum); for (n=3;n12;n+) a1=a1+a2; b1=b1+b2; sum+=b1/a1; printf( %d sum of %fn,n,sum); a2=a2+a1; b2=b2+b1; sum+=b2/a2; printf( %d sum of %fn,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 #include main() int sum=1; int n; printf(The number of peachsis n); for(n=1;n10;n+) sum=(sum+1)*2; printf(The %d day peach is:,10-n); printf(%dn,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 #include main() float a; double x1,x2; printf(Please input the number of an); scanf(%f,&a); if (fabs(a)1e-5) x1=1.0/2.0*(x2+a/x2); printf(x1 %fn,x1); x2=1.0/2.0*(x1+a/x1); printf(x2 %fn,x2); printf(the value of sqrt(a) is :%10.4fn,x1); 5.10打印出以下图案* * * * * * * * * * * * * * * * * * * * * * * *#include #include 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