【水打表】#69 A. Panoramix's Prediction

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#Panoramixs Predictio #Codeforces #Beta Round #69

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在古罗马背景下，通过判断连续两天击败的士兵数量是否构成素数及其紧随的素数，来预测末日是否来临。此问题需遍历50以内的素数列表进行比对。

A prime number is a number which has exactly two distinct divisors: one and itself. For example, numbers 2, 7, 3 are prime, and 1, 6,4 are not.

The next prime number after x is the smallest prime number greater than x. For example, the next prime number after 2 is 3, and the next prime number after 3 is 5. Note that there is exactly one next prime number after each number. So 5 is not the next prime number for 2.

One cold April morning Panoramix predicted that soon Kakofonix will break free from his straitjacket, and this will be a black day for the residents of the Gallic countryside.

Panoramix's prophecy tells that if some day Asterix and Obelix beat exactly x Roman soldiers, where x is a prime number, and next day they beat exactly y Roman soldiers, where y is the next prime number after x, then it's time to wait for Armageddon, for nothing can shut Kakofonix up while he sings his infernal song.

Yesterday the Gauls beat n Roman soldiers and it turned out that the number n was prime! Today their victims were a troop of mRomans (m > n). Determine whether the Gauls should wait for the black day after today's victory of Asterix and Obelix?

Input

The first and only input line contains two positive integers — n and m (2 ≤ n < m ≤ 50). It is guaranteed that n is prime.

Pretests contain all the cases with restrictions 2 ≤ n < m ≤ 4.

Output

Print YES, if m is the next prime number after n, or NO otherwise.

Sample test(s)

input
3 5

output
YES

input
7 11

output
YES

input
7 9

output
NO

问m是不是n的下一个整数，那就先把50以内的素数背出来呗~

然后从第一个开始一个一个往下看~ 啊当然如果m不在这个素数表里那肯定就是NO啦，然后看n是否在m和m前的素数间啦，要注意的是区间范围是左闭右开哦

#include <cstdio>
#include <string>
#include <cstring> 
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
// http://codeforces.com/contest/80  
// Panoramix's Prediction1 

int prime[]={2,3,5,7,11,13,17,19,23,29,31,37,41,43,47};

int main()
{
	int n,m;	cin>>n>>m;
	for(int i=0;i<=14;i++)
	{
		if(prime[i]>n)
		{
			if(prime[i]==m)
			{
				cout<<"YES";
				return 0;
			}
			cout<<"NO";
			return 0;
		}
	}
	cout<<"NO";
	return 0;
}