计蒜客之搜索篇(1)

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1. 回文质数

method 1（512ms）：

/*回文质数
优化:偶数位的回文数整除11,故不为质数
*/
#include <iostream>
#include <algorithm>
#include <cmath>
using namespace std;

int s, e;

void pre() {
    int out[5] = { 2,3,5,7,11 };
    for (int cnt = 0; cnt < 5; cnt++) {
        if (out[cnt] >= s && out[cnt] <= e) {
            cout << out[cnt] << '\n';
        }
    }
    s = max(13, s);
    if (s % 2 == 0) s += 1;
}

bool palin_odd_num(int &i) {
    int out[8];
    int cnt = 0;
    int tmpi = i;
    int mul = 1;
    bool flag = true;
    while (tmpi) {
        out[cnt] = tmpi % 10;
        tmpi /= 10;
        mul *= 10;
        cnt++;
    }
    if (cnt % 2 == 0) {
        i = mul - 1; //当前位数的最后一位
        flag = false;
    }
    else {
        for (int i = 0; i < cnt / 2; i++) {
            if (out[i] != out[cnt - i - 1]) {
                flag = false;
                break;
            }
        }
    }
    return flag;
}

bool prime(int i) { //适用:i>=2
    int level = (int)sqrt(i);
    bool flag = true;
    for (int j = 2; j <= level; j++) {
        if (i % j == 0) {
            flag = false;
            break;
        }
    }
    return flag;
}

void solve() {
    pre();
    for (int i = s; i <= e; i+=2) { //只考虑奇数部分
        if (palin_odd_num(i) && prime(i)) cout << i << '\n';
    }
}

int main() {
    cin >> s >> e;
    solve();
    return 0;
}

method 2（12ms）：

#include <iostream>
#include <algorithm>
#include <cmath>
using namespace std;

int s, e;

void pre() {
    int out[5] = { 2,3,5,7,11 };
    for (int cnt = 0; cnt < 5; cnt++) {
        if (out[cnt] >= s && out[cnt] <= e) {
            cout << out[cnt] << '\n';
        }
    }
}

int getbits(int num) {
    int cnt = 0;
    while (num) {
        num /= 10;
        cnt++;
    }
    return cnt;
}

int getv(int *bits, int bit) {
    int v = 0;
    //[abcd(e)]dcba
    for (int i = 1; i <= bit; i++) {
        v = v * 10 + bits[i];
    }
    for (int i = bit-1; i >= 1; i--) {
        v = v * 10 + bits[i];
    }
    return v;
}

bool prime(int i) { //适用:i>=2
    int level = (int)sqrt(i);
    bool flag = true;
    for (int j = 2; j <= level; j++) {
        if (i % j == 0) {
            flag = false;
            break;
        }
    }
    return flag;
}

void dfs(int i, int tmpbit, int bit, int *bits) {
    if (tmpbit == bit) {
        int v = getv(bits, bit);
        if (v >= s && v <= e && prime(v)) cout << v << '\n';
        return;
    }
    for (int k = 0; k <= 9; k++) { //0~9
        bits[tmpbit+1] = k;
        dfs(k, tmpbit + 1, bit, bits);
    }
}

void solve() {
    pre(); //处理位数为1、2的情况
    int bits[5]; //奇数位最大为7-->(7+1)/2
    int s_bit = getbits(s);
    int e_bit = getbits(e);
    if (s_bit == 1) s_bit = 3; //
    if (e_bit == 9) e_bit = 7;
    for (int bit = (s_bit % 2) ? s_bit : (s_bit + 1); bit <= e_bit; bit+=2) { //去掉偶数位
        for (int i = 1; i <= 9; i += 2) { //首位数字为奇数
            bits[1] = i;
            dfs(i, 1, (bit + 1) / 2, bits);
        }
    }
}

int main() {
    cin >> s >> e;
    solve();
    return 0;
}

method 3：
打表。。。万不得已简单快捷的方法

参照：
【1】回文质数：https://blog.csdn.net/a1033025319/article/details/46969809