标签：codeforces acm
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.
The first line of the input contains two integers n (2?≤?n?≤?1000) and k (0?≤?k?≤?1000).
Output YES if at least k prime numbers from 2 to n inclusively can be expressed as it was described above. Otherwise output NO.
27 2
YES
45 7
NO
In the first sample the answer is YES since at least two numbers can be expressed as it was described (for example, 13 and 19). In the second sample the answer is NO since it is impossible to express 7 prime numbers from 2 to 45 in the desired form.
求n以内等于两个连续素数的和加上1的数的个数 n不大于1000;
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
const int N = 1000;
int n, k, ans;
bool isPrime (int a)
{
for (int i = 2; i <= sqrt (a); ++i)
if (a % i == 0) return 0;
return 1;
}
int main()
{
int ai = 1, bi = 0, a[N], b[N];
a[1] = 2;
for (int i = 3; i <= N; ++i)
if (isPrime (i))
{
a[++ai] = i;
b[++bi] = a[ai] + a[ai - 1] + 1;
}
scanf ("%d%d", &n, &k);
ans = 0;
for (int i = 13; i <= n; ++i)
if (isPrime (i) && find (b + 1, b + bi + 1, i) != b + bi + 1)
ans++;
if (ans >= k) printf ("YES\n");
else printf ("NO\n");
return 0;
}
codeforces 17A Noldbach problem,布布扣,bubuko.com
codeforces 17A Noldbach problem
标签：codeforces acm
原文地址：http://blog.csdn.net/iooden/article/details/38278859
