案 #include <stdio. h> #include <math. h> nt maino Int n,1 printf( Input an integer: scanf( %d", &n) if(n %2==0) printf( %d is an even number. \n", n) else printf( %d is an odd number. \n", n) if(n>1) for(i=2: i<=sgrt(n); i++) f(n%i==0) printf(%d is not a prime \ n", n) goto loop 6
6 案。 #include <stdio.h> #include <math.h> int main() { int n,i; printf("Input an integer: "); scanf("%d",&n); if(n % 2==0) printf("%d is an even number.\n",n); else{ printf("%d is an odd number.\n",n); if(n>1) { for(i = 2; i <= sqrt(n); i++) { if(n%i==0) { printf("%d is not a prime\n",n); goto loop;
printf( %d is a prime \n",n e⊥se loop: return 0 5.7三个成等差数列的数,首尾两项之积为中项的5倍,后两 项之和为第一项的8倍,求此三数。 解:从数学上不难分析它们之间的关系,可编程序如下: #include <stdio. h> #include < math. h> int maino float b =4.0. d. b++:d qr Iwhile(2 * b! =3 d) printf (%.2f \t %. 2f \t %.2f \n", b-d, b, b+d)
7 } } printf("%d is a prime\n",n); } }/* else */ loop: return 0; } 5.7 三个成等差数列的数，首尾两项之积为中项的 5 倍，后两 项之和为第一项的 8 倍，求此三数。 解：从数学上不难分析它们之间的关系，可编程序如下： #include <stdio.h> #include <math.h> int main() { float b = 4.0,d; do{ b++; d = sqrt(b * b – 5 * b); }while(2 * b != 3 * d); printf("%.2f \t %.2f \t %.2f \n",b-d,b,b+d);