Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 28963 Accepted Submission(s): 9629
Problem Description
Give you a lot of positive integers, just to find out how many prime numbers there are.
Input
There are a lot of cases. In each case, there is an integer N representing the number of integers to find. Each integer won’t exceed 32-bit signed integer, and each of them won’t be less than 2.
Output
For each case, print the number of prime numbers you have found out.
Sample Input
3
2 3 4
Sample Output
2
Author
wangye
Source
HDU 2007-11 Programming Contest_WarmUp
問題鏈接:HDU2138 How many prime numbers
問題簡述:統計素數的數量。
問題分析:
簡單模板題,不解釋。
程序說明:(略)
參考鏈接:(略)
題記:(略)
AC的C++語言程序如下:
/* HDU2138 How many prime numbers */
#include <bits/stdc++.h>
using namespace std;
typedef unsigned long long ULL;
const int TIME = 5;
// 快速模冪計算
ULL powmod(ULL a, ULL n, ULL m)
{
ULL res = 1LL;
while(n) {
if(n & 1LL) { // n % 2 == 1
res *= a;
res %= m;
}
a *= a;
a %= m;
n >>= 1;
}
return res;
}
ULL random(ULL n)
{
return (ULL)((double) rand() / RAND_MAX * n + 0.5);
}
bool check(ULL a, ULL n)
{
ULL m = n - 1;
int j = 0;
while((m & 1) == 0) {j++; m >>= 1;}
ULL x = powmod(a, m, n);
if(x == 1 || x == n - 1)
return false;
while(j--) {
x = x * x % n;
if(x == n - 1)
return false;
}
return true;
}
bool Miller_Rabin(ULL n)
{
if(n < 2) return false;
if(n == 2) return true;
if((n & 1) == 0) return false;
ULL x = n - 1;
ULL t = 0;
while((x&1) == 0) {
x >>= 1;
t++;
}
for(int i = 0; i <= TIME; i++)
{
ULL a = random(n - 2) + 1;
if(check(a, n)) return false;
}
return true;
}
int main()
{
int n;
ULL m;
while(~scanf("%d", &n)) {
int ans = 0;
while(n--) {
scanf("%llu", &m);
if(Miller_Rabin(m)) ans++;
}
printf("%d\n", ans);
}
return 0;
}