为什么我这样写函数直接跳出
#include<stdio.h>
#include<math.h>
int z(int i){
for(int o=2;o<=floor(sqrt(i)+0.5);o++){
if(i%o==0)return 0;
}
return 1;
}
int h(int i){
if(i==5&&i==7)return 1;
else if(101<=i&&i<=999&&(i%100==i/10))return 1;
else if(10001<=i&&i<=99999&&(i/10000==i%100000))return 1;
else if(1000001<=i&&i<=9989899&&(i%10000000==i/1000000))return 1;
return 0;
}
int main(){
int a,b;
scanf("%d %d",&a,&b);
for(int i=a;i<=b;i++){
if(z(i)==1&&(h(i)==1)){
printf("%d\n",i);
}
}
return 0;
}