zoukankan      html  css  js  c++  java
  • ZOJ Problem Set

    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.
    题意很简单,就是求长度为n的b进制数在每一位都是1的情况下,是不是素数

    #include <stdio.h>
    #include<math.h>
    int su(long long x);
    int main()
    {
        long long s,p;
        int n,m,i,j;
        while(~scanf("%d%d",&n,&m))
        {
            s=0;
            while(m--)
            {
                s=s+pow(n,m);
            }
                if(su(s)==1)
                {printf("YES
    ");}
            else
                {printf("NO
    ");}
        }
        return 0;
    }
    int su(long long x)
    {
        long long i;
        if(x<2)return 0;
        for(i=2;i*i<=x;i++)
            {
                if(x%i==0)
                  return 0;
            }
        return 1;
    }
    
  • 相关阅读:
    私有程序集的探测过程
    程序集版本控制
    浅谈对对象clone的理解
    [导入]WCF后传系列(3):深入WCF寻址Part 3—消息过滤引擎
    [导入]WCF后传系列(5):深入WCF寻址Part 5—逻辑地址和物理地址
    强名称程序集
    [导入]WCF后传系列(4):深入WCF寻址Part 4—自定义消息筛选器
    绑定过程小结
    概述
    UpdatePanel 的更新与触发环境
  • 原文地址:https://www.cnblogs.com/webmen/p/5739716.html
Copyright © 2011-2022 走看看