ZOJ - 3758 - Singles' Day（素数判定）

Singles’ Day

Time Limit: 2 Seconds    Memory Limit: 65536 KB

Singles’ Day(or One’s Day), an unofficial holiday in China, is a pop culture entertaining holiday on November 11 for young Chinese to celebrate their bachelor life. With the meaning of single or bachelor of number ‘1’ and the huge population of young single man. This festival is very popular among young Chinese people. And many Young bachelors organize parties and Karaoke to meet new friends or to try their fortunes that day.

On Singles’ Day, a supermarket has a promotional activity. Each customer will get a ticket on which there are two integers b and N, representing an integer M that only contains N digits 1 using b as the radix. And if the number M is a prime number, you will get a gift from the supermarket.

Since there are so many customers, the supermarket manager needs your help.

Input

There are multiple test cases. Each line has two integers b and N indicating the integer M, which might be very large. (2 <= b <= 16, 1 <= N <= 16)

Output

If the customer can get a gift, output “YES”, otherwise “NO”.

Sample Input

3 3
2 4
2 1
10 2

Sample Output

YES
NO
NO
YES

Hint

For the first sample, b=3, N=3, so M=(111)3, which is 13 in decimal. And since 13 is a prime number, the customer can get a gift, you should output “YES” on a line.

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题意：给你一组b， n，代表有n个1的b进制，判断它对应的十进制数是否是一个素数

思路：一看到素数判定，第一反应就是素数筛选法打表，再一看范围，我傻了，最大是16个1的16进制，转换了一下，这也太大了吧。。。 算是get了一个新技能，快速素数判定法，但我依旧喜欢素数筛选，范围不大的话还是筛选法好用

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;

long long pow_n(int n, int k)
{
    long long sum = 1;
    if(!n)
        return 1;
    for(int i = 1; i<=n; i++)
        sum*=k;
    return sum;
}

bool prime (long long num)
{
    if (num == 2 || num == 3 || num == 5)
        return true;
    if (num % 2 == 0 || num % 3 == 0 || num % 5 == 0 || num == 1)
        return false;

    long long c = 7;
    int maxc = (int)(sqrt (num));
    while (c <= maxc)
    {
        if (num % c == 0)
            return false;
        c += 4;
        if (num % c == 0)
            return false;
        c += 2;
        if (num % c == 0)
            return false;
        c += 4;
        if (num % c == 0)
            return false;
        c += 2;
        if (num % c == 0)
            return false;
        c += 4;
        if (num % c == 0)
            return false;
        c += 6;
        if (num % c == 0)
            return false;
        c += 2;
        if (num % c == 0)
            return false;
        c += 6;
    }
    return true;
}


int main()
{
    int n, b;
    while(scanf("%d %d", &b, &n)!=EOF)
    {
        long long m = 0;               //注意范围，longlong就足够了
        for(int i = 0; i < n; i++)     //把b进制转换为10进制
            m += pow_n(i,b);

    bool flag = prime(m);
    if(!flag)
            printf("NO\n");
    else
        printf("YES\n");

    }

    return 0;
}