题解 | #对称平方数1#

#include <iostream>
#include "stdio.h"
bool duichen(int n2) {
    int ge, shi, bai, qian, wan;
    ge = n2 % 10;
    shi = (n2 / 10) % 10;
    bai = (n2 / 100) % 10;
    qian = (n2 / 1000) % 10;
    wan = n2 / 10000;
    if (n2 < 10)
        return true;
    if (n2 > 10 && n2 < 100 && ge == shi) {
        return true;
    }
    if (n2 > 100 && n2 < 1000 && ge == bai) {
        return true;
    }
    if (n2 > 1000 && n2 < 10000 && ge == qian && shi == bai) {
        return true;
    }
    if (n2 > 10000 && n2 < 100000 && ge == wan && shi == qian) {
        return true;
    }

    return false;
}

int main() {
    int n, n2;
    for (n = 0; n <= 256; n++) {
        n2 = n * n;
        if (duichen(n2))
            printf("%d\n", n);
    }
}