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HDOJ 5150 Sum Sum Sum Miller_Rabin

时间:2014-12-27 23:11:06      阅读:207      评论:0      收藏:0      [点我收藏+]

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很少有这么裸的题目,测一下Miller_Rabin


Sum Sum Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 72    Accepted Submission(s): 52


Problem Description
We call a positive number X P-number if there is not a positive number that is less than X and the greatest common divisor of these two numbers is bigger than 1.
Now you are given a sequence of integers. You task is to calculate the sum of P-numbers of the sequence.
 

Input
There are several test cases. 
In each test case:
The first line contains a integer N(1N1000). The second line contains N integers. Each integer is between 1 and 1000.
 

Output
For each test case, output the sum of P-numbers of the sequence.
 

Sample Input
3 5 6 7 1 10
 

Sample Output
12 0
 

Source
 



/* ***********************************************
Author        :CKboss
Created Time  :2014年12月27日 星期六 21时51分17秒
File Name     :HDOJ5150.cpp
************************************************ */

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>

using namespace std;

/// Miller_Rabin

typedef long long int LL;

const int S=8;///判断几次 (8~12)

/// (a*b)%c
LL mult_mod(LL a,LL b,LL c)
{
	a%=c; b%=c;
	LL ret=0; LL temp=a;
	while(b)
	{
		if(b&1)
		{
			ret+=temp;
			if(ret>c) ret-=c;
		}
		temp<<=1;
		if(temp>c) temp-=c;
		b>>=1LL;
	}
	return ret;
}

/// (a^n)%mod
LL pow_mod(LL a,LL n,LL mod)
{
	LL ret=1;
	LL temp=a%mod;
	while(n)
	{
		if(n&1) ret=mult_mod(ret,temp,mod);
		temp=mult_mod(temp,temp,mod);
		n>>=1LL;
	}
	return ret;
}

/// check a^(n-1)==1(mod n)
bool check(LL a,LL n,LL x,LL t)
{
	LL ret=pow_mod(a,x,n);
	LL last=ret;
	for(int i=1;i<=t;i++)
	{
		ret=mult_mod(ret,ret,n);
		if(ret==1&&last!=1&&last!=n-1) return true;
		last=ret;
	}
	if(ret!=1) return true;
	return false;
}

bool Miller_Rabin(LL n)
{
	if(n<2) return false;
	if(n==2) return true;
	if((n&1)==0) return false;
	LL x=n-1;
	LL t=0;
	while((x&1)==0) { x>>=1; t++;}
	srand(time(NULL));

	for(int i=0;i<S;i++)
	{
		LL a=rand()%(n-1)+1;
		if(check(a,n,x,t)) return false;
	}
	return true;
}

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);

	int n;
	while(scanf("%d",&n)!=EOF)
	{
		LL sum=0; LL x;
		for(int i=0;i<n;i++)
		{
			cin>>x;
			if(Miller_Rabin(x)||x==1) 
			{
				sum+=x;
			}
		}
		cout<<sum<<endl;
	}
    
    return 0;
}


HDOJ 5150 Sum Sum Sum Miller_Rabin

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原文地址:http://blog.csdn.net/ck_boss/article/details/42200103

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