2015
04-13

Twin Prime Conjecture

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.

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.

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.

1
5
20
-2

0
1
4

#include <stdio.h>
#include <math.h>
int a[10000]={2},num[100000]={0,0};
int main()
{
int n,i,j,g,temp;
for(g=0,i=3;i<100000;i++)
{
temp=sqrt(i);
for(j=2;j<=temp;j++)
if(i%j==0)break;
if(j>temp)
{
a[g]=i;
num[i]=num[i-1]+(a[g-1]+2==a[g]?1:0);
g++;
}
else
num[i]=num[i-1];
}
while(scanf("%d",&n)&&n>=0)
{
printf("%d\n",num[n]);
}
return 0;
}

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2. 在方法1里面：

//遍历所有的边，计算入度
for(int i=0; i<V; i++)
{
degree = 0;
for (j = adj .begin(); j != adj .end(); ++j)
{
degree[*j]++;
}
}

为什么每遍历一条链表，要首先将每个链表头的顶点的入度置为0呢？
比如顶点5，若在顶点1、2、3、4的链表中出现过顶点5，那么要增加顶点5的入度，但是在遍历顶点5的链表时，又将顶点5的入度置为0了，那之前的从顶点1234到顶点5的边不是都没了吗？

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