HDU 3792Twin Prime Conjecture

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题解：首先打表找出素数，并且找到相邻的两个素数相差为2。接着用树状数组更新数据。

ac code：

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <cstdlib>#include <cmath>#include <vector>#include <list>#include <deque>#include <queue>#include <iterator>#include <stack>#include <map>#include <set>#include <algorithm>#include <cctype>using namespace std;#define si1(a) scanf("%d",&a)#define si2(a,b) scanf("%d%d",&a,&b)#define sd1(a) scanf("%lf",&a)#define sd2(a,b) scanf("%lf%lf",&a,&b)#define ss1(s)  scanf("%s",s)#define pi1(a)    printf("%d\n",a)#define pi2(a,b)  printf("%d %d\n",a,b)#define mset(a,b)   memset(a,b,sizeof(a))#define forb(i,a,b)   for(int i=a;i<b;i++)#define ford(i,a,b)   for(int i=a;i<=b;i++)typedef long long LL;const int N=1100001;const int M=6666666;const int INF=0x3f3f3f3f;const double PI=acos(-1.0);const double eps=1e-7;#define maxn 100006#define lowbit(x) x&(-x)int prime[maxn]={0},vis[maxn],num=0,b[maxn],n;int sum(int x){    int ans=0;    while(x>0)    {        ans+=b[x];        x-=lowbit(x);    }    return ans;}void update(int x){    while(x<=maxn)    {        b[x]++;        x+=lowbit(x);    }}int main(){    vis[0]=1,vis[1]=1;    for(int i=2;i<maxn;i++)    {        if(!vis[i])prime[num++]=i;        for(int j=0;j<num&&i*prime[j]<maxn;j++)        {            vis[i*prime[j]]=1;            if(!(i%prime[j]))break;        }    }//先打表    for(int i=1;i<maxn;i++)    {        if(prime[i]-prime[i-1]==2)        {            update(prime[i]);//更新树状数组        }    }    //for(int i=0;i<=20;i++)        //printf("%d\n",b[i]);    while(~si1(n)&&n>=0)    {        printf("%d\n",sum(n));//查找前面符合条件的值    }    return 0;}

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