HDU-2973-YAPTCHA（威尔逊定理）

YAPTCHA

Problem Description

The math department has been having problems lately. Due to immense amount of unsolicited automated programs which were crawling across their pages, they decided to put Yet-Another-Public-Turing-Test-to-Tell-Computers-and-Humans-Apart on their webpages. In short, to get access to their scientific papers, one have to prove yourself eligible and worthy, i.e. solve a mathematic riddle.

However, the test turned out difficult for some math PhD students and even for some professors. Therefore, the math department wants to write a helper program which solves this task (it is not irrational, as they are going to make money on selling the program).

The task that is presented to anyone visiting the start page of the math department is as follows: given a natural n, compute

where [x] denotes the largest integer not greater than x.

Input

The first line contains the number of queries t (t <= 10⁶). Each query consist of one natural number n (1 <= n <= 10⁶).

Output

For each n given in the input output the value of Sn.

Sample Input

13
1
2
3
4
5
6
7
8
9
10
100
1000
10000
 

Sample Output

0
1
1
2
2
2
2
3
3
4
28
207
1609

解题思路：

威尔逊定理：
(p−1)! ≡ −1(mod p) (p 为素数)
那么上图中可以发现，当p为素数是，表达式为1 反之为0。
那么问题就变成了求k以内 3k+7为素数的个数。
打表判断。 前缀和优化复杂度。

AC代码：

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
#define sd(n) scanf("%d",&n)
#define sdd(n,m) scanf("%d%d",&n,&m)
#define sddd(n,m,k) scanf("%d%d%d",&n,&m,&k)
#define pd(n) printf("%d\n", n)
#define pc(n) printf("%c", n)
#define pdd(n,m) printf("%d %d", n, m)
#define pld(n) printf("%lld\n", n)
#define pldd(n,m) printf("%lld %lld\n", n, m)
#define sld(n) scanf("%lld",&n)
#define sldd(n,m) scanf("%lld%lld",&n,&m)
#define slddd(n,m,k) scanf("%lld%lld%lld",&n,&m,&k)
#define sf(n) scanf("%lf",&n)
#define sc(n) scanf("%c",&n)
#define sff(n,m) scanf("%lf%lf",&n,&m)
#define sfff(n,m,k) scanf("%lf%lf%lf",&n,&m,&k)
#define ss(str) scanf("%s",str)
#define rep(i,a,n) for(int i=a;i<=n;i++)
#define per(i,a,n) for(int i=n;i>=a;i--)
#define mem(a,n) memset(a, n, sizeof(a))
#define debug(x) cout << #x << ": " << x << endl
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define mod(x) ((x)%MOD)
#define gcd(a,b) __gcd(a,b)
#define lowbit(x) (x&-x)
#define pii map
#define mk make_pair
#define rtl rt<<1
#define rtr rt<<1|1
#define Max(x,y) (x)>(y)?(x):(y)
#define int long long


typedef pair<int,int> PII;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
const int MOD = 1e9 + 7;
const ll mod = 10007;
const double eps = 1e-9;
const ll INF = 0x3f3f3f3f3f3f3f3fll;
//const int inf = 0x3f3f3f3f;
inline int read(){int ret = 0, sgn = 1;char ch = getchar();
while(ch < '0' || ch > '9'){if(ch == '-')sgn = -1;ch = getchar();}
while (ch >= '0' && ch <= '9'){ret = ret*10 + ch - '0';ch = getchar();}
return ret*sgn;}
inline void Out(int a){if(a>9) Out(a/10);putchar(a%10+'0');}
int qpow(int m, int k, int mod){int res=1%mod,t=m%mod;while(k){if(k&1)res=res*t%mod;t=t*t%mod;k>>=1;}return res;}
ll gcd(ll a,ll b){if(b > a) swap(a,b); return b==0?a : gcd(b,a%b);}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll inv(ll x,ll mod){return qpow(x,mod-2,mod)%mod;}

const int N = 3e6+15;
signed prime[N],mark[N],pcnt;

// 如果变量名都相同的话,就不用传参了
//void getPrimes(int prime[],int N,int &pcnt)
void getPrimes()
{
    memset(mark,0,sizeof(mark));
    mark[0] = mark[1] = 1;
    pcnt = 0;
    for(int i = 2; i < N ; i ++)
    {
        if(!mark[i])
        	prime[pcnt++] = i;
        for(int j = 0 ; (ll)i*prime[j] < N && j < pcnt ; j ++)
        {
            mark[i*prime[j]] = 1;
            if(i%prime[j] == 0)
                break;
        }
    }
}

signed res[N];

signed main()
{
    getPrimes();
    int t = 1,cas = 1;
    //cin>>t;
    for(int i = 0 ; i < (int)1e6+7 ; i ++)
    {
        if(!mark[3*i+7])
            res[i] = 1;
        else
            res[i] = 0;
    }
    for(int i = 2 ; i < (int)1e6+7 ; i ++)
        res[i] += res[i-1];
    cin>>t;
    while(t--)
    {
        int n;
        cin>>n;
        cout<<res[n]<<endl;

    }
}