prime palindrome

The number 151 is a prime palindrome because it is both a prime number and a palindrome (it is the same number when read forward as backward). Write a program that finds all prime palindromes in the range of two supplied numbers a and b (5 <= a < b <= 1000,000,000); both a and b are considered to be within the range .

素数回文数判定，数据范围大，对时间要求较为严苛。

下以8位数为例，证明偶数位数的回文数都不是prime number 

转自liujiacongBaBa https://blog.csdn.net/jc514984625/article/details/53461821

设x=10000001i+1000010j+100100k+11000l(i为1-9的自然数,j,k,l为0-9的自然数)

分解得x=11(909091i+90910j+9100k+1000l)

又909091i+90910j+9100k+1000l>2

∴11|x

x必定为合数

同理可得当回文数x的位数为偶数位时必有11|x当x/11>2时x必为合数

 直接枚举回文数的前几位，使用mir函数计算整个数字（如果枚举后几位可能会在末尾的0处出现问题），使用is_prime判断是否素数并输出

#include <stdio.h>
#include <string.h>
#include <math.h>
int a, b;
int x;
int ans;
int t;

int is_prime(int p) {
    if (p == 1) return 0;
    int x2 = sqrt(p) + 1;
    for (int i = 2; i <= x2; i++) {
        if (p % i == 0) return 0;
    }
    return 1;
}

int mir(int ori, int base, int po) {
    ans = 0;
    t = ori / 10;
    for (int i = 1; i <= base; i++) {
        ans *= 10;
        ans += (t % 10);
        t /= 10;
    }
    return (ans + ori * po);
}

void check(int base) {
    int tcheck = 1;
    for (int i = 1; i <= 2 * base - 2; i++) tcheck *= 10;
    if (tcheck > b) return;
    tcheck = 9;
    for (int i = 1; i <= 2 * base - 2; i++) {
        tcheck *= 10;
        tcheck += 9;
    }
    if (tcheck < a) return;
    int imin = 1;
    for (int i = 1; i < base; i++) imin *= 10;
    // if (imin > b) return;
    int imax = 9;
    for (int i = 1; i < base; i++) {
        imax *= 10;
        imax += 9;
    } 
    for (int i = imin; i <= imax; i++) {
        x = mir(i, base - 1, imin);
        if (x == 9) x = 11;
        if (x < a || x > b) continue;
        if(is_prime(x)) printf("%d\n", x);
    }
    // if (imax < a) return;
    // printf("%d %d %d\n", base, imin, imax);
}  
int main () {
    scanf("%d %d", &a, &b);
    for (int i = 1; i <= 5; i++) {
        // if (i % 2 == 0) continue;
        check(i);
    }
    // printf("%d", mir(15550, 4, 10000));

    return 0;
}