HDU 2973 YAPTCHA（威爾遜定理）

Mr_Treeeee發表於2020-04-06
APT

YAPTCHA

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 930    Accepted Submission(s): 491


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 <= 10^6). Each query consist of one natural number n (1 <= n <= 10^6).
 

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





 



Source



Central European Programming Contest 2008 



 





題意：


給了你一個公式，讓你求這個公式。





POINT：


利用威爾遜定理：


若p為質數，則p可整除(p-1)!+1。




若3k+7為質數，就是1，其他情況則是0。


實求x屬於（1，n）時3*x+7中有幾個質數。





#include <iostream>
#include <string.h>
#include <stdio.h>
#include <stack>
#include <algorithm>
#include <math.h>
using namespace std;
#define ll long long
//int a[1000000];
int num=0;
int f[3000000+10];
int ans[1000000];
void init()
{
    memset(f,1,sizeof f);
   // a[++num]=2;
    for(int i=4;i<=3000007;i+=2) f[i]=0;
    for(int i=3;i<=3000007;i++)
    {
        if(f[i])
        {
           // a[++num]=i;
            for(int j=i*3;j<=3000007;j=j+i*2)
            {
                f[j]=0;
            }
        }
    }
    ans[0]=0;
    for(int i=1;i<=1e6;i++)
    {
        if(f[i*3+7])
        {
            ans[i]=ans[i-1]+1;
        }
        else
            ans[i]=ans[i-1];
    }
  //  printf("%d",num);
}
int main()
{
    init();
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int n;
        scanf("%d",&n);
        printf("%d\n",ans[n]);
    }
    
}






















                                    




相關文章