printf("Computer take %d stacks." s) if ((total-s)<=0) i printf(" *kk You Win! Game Over. *k*k\n") else total=total-s I s=total%5-i-l f(s=-1)s=4 if (s if(s==-4)s=1 total=total-i-s printf(" Computer take %d stacks." s) printf(" %d sticks left. \n, total 10.5求不超过1000的回文素数。 解:关键点是构造回文整数和判断素数,不超过1000的回文数包括两位和三位数,这里用 穷举法构造一个整数,并求出与其对应的反序数,两者相同既是回文数。参 考程序如下: #include <stdio. h> int prime(int n) I int i for(i=2;i<n/2;i++) if(n%i==0) return 0 return 1 maino k, s, t: printf(follow is output: \n") for(i=0;i<=9;i+) for(j=0;j<=9;j+) for(k=0;k<=9;k+) s=i*100+j*10+kprintf(" Computer take %d stacks. ",s); if ((total-s)<=0) { printf(" *** You Win! Game Over. ***\n"); break; } else total=total-s; } else { s=total%5-i-1; if (s==-1) s=4; if (s==-2) s=3; if (s==-3) s=2; if (s==-4) s=1; total=total-i-s; printf(" Computer take %d stacks. ",s); } printf(" %d sticks left.\n",total); } } 10.5 求不超过 1000 的回文素数。 解：关键点是构造回文整数和判断素数，不超过 1000 的回文数包括两位和三位数，这里用 穷举法构造一个整数，并求出与其对应的反序数，两者相同既是回文数。参 考程序如下： #include <stdio.h> int prime(int n) { int i; for (i=2;i<n/2;i++) if (n%i==0) return 0; return 1; } main() {int i,j,k,s,t; printf("Follow is output:\n"); for (i=0;i<=9;i++) for (j=0;j<=9;j++) for (k=0;k<=9;k++) { s=i*100+j*10+k;