int a=2,n=5,s; s=a; for(;--n;) s=s*10+a; printf("%d",s);
int f(int n); main() { int a=3,s; s=f(a); s=s+f(a); printf("
int f(int n); main() { int a=3,s; s=f(a); s=s+f(a); printf(&
char s[]=" an apple" ; printf(" %d\n",strlen(s)); A)7 B)8 C)
(i!=p) 求解释main(){ int i,j,p,q,s,a[10]; printf("\n input 10 n
#define S(x)4 *(x)*x+1 main() {int k=5,j=2;printf("%d\n",S(k
main() { int a,s,max; scanf("%d%d",&a,&s); if(a>s) { max=a }
#include int f(int n); main() {int a=3,s; s=f(a);s=s+=(a);pr
printf("%s%s%s%d%d%d\n",printf("\n");
#include main() { float a,n,i,j,s; scanf("%d",&a); for(n=1.0
# include # include int mian () { int m,n,i,s=0; int a[10000
main() { char s[]="stop\0\n\""; printf("%d\n",strlen(s)) }
VFP中 CLEAR DIMENSION a(3) FOR N=1 TO 3 S=4 a(N)=N*2+1 S=S+a(