C语言程序设计 谭浩强 习题答案.doc

C语言程序设计 谭浩强 习题答案与大家分享5.7#includestdio.hvoid main()int x,a,b,c,d,e,num;scanf(%d,&x);a=(x-x%10000)/10000;b=(x-10000*a-(x-10000*a)%1000)/1000;c=(x-x%100)/100-a*100-b*10;d=(x-x%10)/10-a*1000-b*100-c*10;e=x%10;if(x/100000) num=5;else if(x/10000) num=4;else if(x/1000) num=3;else if(x/100) num=2;else num=1;printf(它是%d位数n,num);switch(num)case 1:printf(每一位数字%d;逆序为%dn,e);break;case 2:printf(每一位数字%d,%d;逆序为%d%dn,d,e,e,d);break;case 3:printf(每一位数字%d,%d,%d;逆序为%d%d%dn,c,d,e,e,d,c);break;case 4:printf(每一位数字%d,%d,%d,%d;逆序为%d%d%d%dn,b,c,d,e,e,d,c,b);break;case 5:printf(每一位数字%d,%d,%d,%d,%d;逆序为%d%d%d%d%dn,a,b,c,d,e,e,d,c,b,a);break;default:printf(error);6.1#includestdio.hvoid main()int m,n,a,b,c;scanf(%d,%d,&m,&n);a=m%n;b=m*n;while(a!=0)m=n;n=a;a=m%n;printf(%d是其最大公约数。,n);c=b/n;printf(%d是其最小公倍数。,c);6.2#includestdio.hvoid main()char a;int b=0,c=0,d=0,e=0;while(a=getchar()!=n)if(a=a&a=A&a=0&a=9)d=d+1;else e=e+1;printf(英文字母个数为%d,空格字数为%d,数字个数为%d,其他字符个数为%d。n,b,c,d,e);6.3#includestdio.hvoid main()int a,m,n;long b=1,sum=0,c;scanf(%d,%d,&a,&n);for(m=1;m=n;m+)sum=sum+b;b=b*10+1;c=sum*a;printf(%ld,c);6.4#includestdio.hvoid main()int i,j;long a=1,sum=0;for(i=1;i=20;i+)for(j=1;j=i;j+)a=a*j;sum=sum+a;printf(%ldn,sum);6.5#includestdio.hvoid main()int i,j,k;long a=0,b=0;float sum,c=0;for(i=1;i=100;i+)a=a+i;for(j=1;j=50;j+)b=b+j*j;for(k=1;k=10;k+)c=c+1.0/k;sum=a+b+c;printf(%fn,sum);6.6#includestdio.hvoid main()int i,j,k,m,n;for(i=1;i10;i+)for(j=1;j10;j+)for(k=1;k10;k+)m=i*i*i+j*j*j+k*k*k;n=i*100+j*10+k;if(m=n)printf(%d是一个水仙花数。n,m);7.1#includestdio.h#includemath.hvoid main()int i,j,t,n=0;for(i=1;i=100;i=i+2)t=sqrt(i);for(j=2;jt)n=n+1;printf(%5d,i);if(n%5=0) printf(n);printf(n);7.2#includestdio.hvoid main()int a10,i,j,t,max;for(i=0;i10;i+)scanf(%d,&ai);for(j=0;j10;j+)max=aj;for(i=j+1;imax)t=aj;aj=ai;ai=t;max=aj;printf(%4d,aj);printf(n);或者#includestdio.hvoid main()int a10,i,j,t;for(i=0;i10;i+)scanf(%d,&ai);for(j=0;j10;j+)for(i=j+1;iaj)t=aj;aj=ai;ai=t;printf(%4d,aj);printf(n);7.3#includestdio.hvoid main()int a33,i,j,sum2=0;for(i=0;i3;i+)for(j=0;j3;j+)scanf(%d,&aij);for(i=0;i3;i+)sum0=sum0+aii;sum1=sum1+ai2-i;printf(主对角线之和为%d，副对角线之和为%d,sum0,sum1);7.4#includestdio.hvoid main()int a10,i,m,k,b11;for(i=0;ia9)for(i=0;i10;i+)bi=ai;b10=m;else if(ma0)for(i=0;i10;i+)bi+1=ai