How to solve the problem of ## includo #?

OJ indicates that the time is out of limit, can you help me change it?

Description
Because 151 is both a prime number and a palindrome number(it looks the same from left to right as from right to left), 151 is a palindrome prime Href = "https://ask.csdn.net/5%3C=a%3Cb%3C=100, 000000" target = "_blank" > a, b < / a > all palindrome between prime Numbers;

Enter
Line 1: Two integers a and b.
Output
Output a list of palindrome primes, one in line.
Sample input
5 500 the sample output 5 7 101 151 131 181 191 353 313 373 383

#include <bits/stdc++.h>
using namespace std;
bool isSymm(int a) {
    int  j, k=0;
    j = a;
    while (j > 0) {
        k = k*10+ j % 10 ;
        j /= 10;
    }
    if (k == a)return true;
    else return false;
}
int main()
{
    int a,b,i=2;
    cin>>a>>b;
    for(;a<=b;a++){
        while(a%i!=0 && i<=a-1)    
        {
            i++;
        }
        if (i>a-1 && isSymm(a))
        {
            cout<<a<<endl;
        }    
        i=2;
    }
    return 0;
}

How to solve the problem of ## includo #?

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0 Answer

cyj773 注册会员
2023-02-27 23:13

OJ or prompt time error ah, please tell me how to save time?

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dapi890621 注册会员
2023-02-27 23:13

See c language - # include < / a > < / li >

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dongwei1529 注册会员
2023-02-27 23:13


#include 
using namespace std;

// 判断是否为回文数
bool isSymm(int a) {
    int j = a, k = 0;
    while (j > 0) {
        k = k * 10 + j % 10;
        j /= 10;
    }
    return k == a;
}

// 判断是否为质数
bool isPrime(int a) {
    if (a == 1) return false;
    if (a == 2) return true;
    for (int i = 2; i <= sqrt(a); i++) {
        if (a % i == 0) {
            return false;
        }
    }
    return true;
}

int main() {
    int a, b;
    cin >> a >> b;
    for (int i = a; i <= b; i++) {
        if (isSymm(i) && isPrime(i)) {
            cout << i << endl;
        }
    }
    return 0;
}

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: daijianlin 注册会员
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Publish Time: 2023-02-27 23:13

Update Time: 2023-02-27 23:13