HDOJ 题目 3792 Twin Prime Conjecture（数学）

Twin Prime Conjecture

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 2249    Accepted Submission(s): 720


Problem Description

If we define dn as: dn = pn+1-pn, where pi is the i-th prime. It is easy to see that d1 = 1 and dn=even for n>1. Twin Prime Conjecture states that "There are infinite consecutive primes differing by 2".

Now given any positive integer N (< 10^5), you are supposed to count the number of twin primes which are no greater than N.

 

Input

Your program must read test cases from standard input.

The input file consists of several test cases. Each case occupies a line which contains one integer N. The input is finished by a negative N.

 

Output

For each test case, your program must output to standard output. Print in one line the number of twin primes which are no greater than N.

 

Sample Input

1
5
20
-2

 

Sample Output

0
1
4

 

Source

浙大计算机研究生保研复试上机考试-2011年

 

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题意
twins prime：素数b-素数a==2，

求0-n有多少twins prime

ac代码

#include<stdio.h>
#include<string.h>
#include<math.h>
int a[100010],b[100010],cot;
void fun()
{
int i,j,w;
cot=1;
a[0]=2;
for(i=3;i<100010;i++)
{
w=1;
for(j=2;j<=sqrt(i);j++)
{

if(i%j==0)
{
w=0;
break;
}
}
if(w)
a[cot++]=i;
}
}
int main()
{
int i,j,k,t,n,ans=0;
fun();
t=2;
for(i=1;i<cot;i++)
{
//b[i]=b[i-1];
if(a[i]-t==2)
ans++;
t=a[i];
b[a[i]]=ans;
}
j=0;
for(i=0;i<100010;i++)
{
if(b[i]>j)
j=b[i];
b[i]=j;
}
while(scanf("%d",&n)!=EOF,n>=0)
{
printf("%d\n",b
);
}
}