Noldbach problem

Description

Noldbach problem

time limit per test: 2 seconds

memory limit per test: 64 megabytes

input: standard input

output: standard output

Nick is interested in prime numbers. Once he read about Goldbach problem. It states that every even integer greater than 2 can be expressed as the sum of two primes. That got Nick's attention and he decided to invent a problem of his own and call it Noldbach problem. Since Nick is interested only in prime numbers, Noldbach problem states that at least k prime numbers from 2 to n inclusively can be expressed as the sum of three integer numbers: two neighboring prime numbers and 1. For example, 19 = 7 + 11 + 1, or 13 = 5+ 7 + 1.

Two prime numbers are called neighboring if there are no other prime numbers between them.

You are to help Nick, and find out if he is right or wrong.

Input

The first line of the input contains two integers n (2 ≤ n ≤ 1000) and k (0 ≤ k ≤ 1000).

Output

Output YES if at least k prime numbers from 2 to n inclusively can be expressed as it was described above. Otherwise output NO.

Sample test(s)

input

27 2

output

YES

 

 

 

 

 

 

input

45 7

output

NO

 

 

 

 

 

 

题解：

　　如果一个素数可以用三个素数之和表示，其中一个为1，另外两个为相邻的素数，则计数一次。

　　从2到n，存在上述的素数不少于k个，输出YES，否则输出NO。

代码：

#include <iostream>
#include <cstdio>
#include <cmath>

int isPrime(int n); //判断n是否是素数
void initPrime(int n);  //将n以内的素数存到数组内 此处用1000即可

using namespace std;
int p[200]; //存储2~1000之间的素数
int main()
{
    int n, k, i, j, counter=0;
    cin >> n >> k;
    initPrime(1000);
    for(i=2; i<=n; i++) {
        if(isPrime(i)==0)
            continue;
        for(j=0; p[j]<i; j++) {
            if(p[j]+p[j+1]+1 == i){
                counter++;
                break;
            }
        }
    }
    if(counter < k)
        cout << "NO" <<endl;
    else
        cout << "YES" <<endl;
    return 0;
}

void initPrime(int n) {
    int i, j = 0;
    for(i=2; i<n; i++) {
        if(isPrime(i)){
            p[j++] = i;
        }
    }
}

int isPrime(int n) {
    int i;
    for(i=2; i<=(int)sqrt(n); i++) {
        if(n%i == 0) {
            return 0;
        }
    }
    return 1;
}

Problem -17A - Codeforces

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