hdu 2973 威尔逊定理

初等数论四大定理之一: 威尔逊定理：当且仅当p为素数时，(p-1)!===-1(modp)..

换句话说：当p为素数时，那么p能整除（p-1）!+1；

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2973

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

1312345678910100100010000

 

Sample Output

0112222334282071609

 

思路：若干k*3+7是素数，则结果为1，不是素数结果为0，所以可以求出1*3+7~n*3+7中的素数个数；

因为询问次数比较多，如果每次都求一遍会TLE，预先打个表处理一下就OK了

#include <iostream>#include <stdio.h>#include <string.h>#include <string>#include <cmath>const int N=3000100;using namespace std;bool isprime[N];//表示第i个数是不是素数int cnt[N];//存储第i个数的最终结果void prime(){ memset(isprime,true,sizeof(isprime)); for(int i=2;i<N;i++) {   if(isprime[i])   for(int j=i+i;j<N;j+=i)   isprime[j]=false; }}int sa(int k){     if(isprime[k*3+7])return 1; return 0;}void eg() //打表处理{ memset(cnt,0,sizeof(cnt)); for(int i=1;i<1000100;i++)  //耗时大概700ms   cnt[i]=cnt[i-1]+sa(i);}int main(){ int T,n; cin>>T; prime(); eg(); while(T--) {  scanf("%d",&n);  printf("%d\n",cnt[n]); } return 0;}