POJ 3006 Dirichlet's Theorem on Arithmetic Progressions 素数难度:0

http://poj.org/problem?id=3006

#include <cstdio>
using namespace std;
bool pm[1000002];
bool usd[1000002];
bool judge(int x)
{
    if(usd[x])return pm[x];
    usd[x] = true;
    if(x == 2) return pm[x] = true;
    if(((x & 1) == 0) || (x < 2))return pm[x] = false;
    for(int i = 3;i * i <= x;i+= 2)
    {
        if(x % i == 0)return pm[x] = false;
    }
    return pm[x] = true;
}

int main(){
    int a,d,n;
    while(scanf("%d%d%d",&a,&d,&n) == 3 && a)
    {
        int cnt = 0;
        for(int tn = 0;tn < n;cnt++)
        {
            if(judge(cnt * d + a))tn++;
            if(tn == n)break;
        }
        printf("%d\n",cnt * d + a);
    }

    return 0;
}

时间： 2024-12-06 18:10:18

POJ 3006 Dirichlet's Theorem on Arithmetic Progressions 素数难度:0的相关文章

poj 3006 Dirichlet's Theorem on Arithmetic Progressions

Description If a and d are relatively prime positive integers, the arithmetic sequence beginning with a and increasing by d, i.e., a, a + d, a + 2d, a + 3d, a + 4d, ..., contains infinitely many prime numbers. This fact is known as Dirichlet's Theore

POJ 3006 Dirichlet's Theorem on Arithmetic Progressions 快筛质数

题目大意:给出一个等差数列,问这个等差数列的第n个素数是什么. 思路:这题主要考如何筛素数,线性筛.详见代码. CODE: #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #define MAX 1000010 using namespace std; int prime[MAX],primes; bool notp[MAX]; int a,d,n;

POJ - 3006 - Dirichlet's Theorem on Arithmetic Progressions = 水题

http://poj.org/problem?id=3006 给一个等差数列,求其中的第n个质数,答案保证不超过1e6.n还特别小?!!! 埃筛之后暴力. #include<algorithm> #include<cmath> #include<cstdio> #include<cstring> #include<iostream> #include<map> #include<set> #include<stack

Dirichlet's Theorem on Arithmetic Progressions POJ - 3006 线性欧拉筛

题意给出a d n 给出数列 a,a+d,a+2d,a+3d......a+kd 问第n个数是几保证答案不溢出直接线性筛模拟即可 1 #include<cstdio> 2 #include<cstring> 3 using namespace std; 4 bool Is_Primes[1000005]; 5 int Primes[1000005]; 6 int A[1000005]; 7 int cnt; 8 void Prime(int n){ 9 cnt=0; 1

（素数求解）I - Dirichlet's Theorem on Arithmetic Progressions(1.5.5)

Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u Submit Status Description If a and d are relatively prime positive integers, the arithmetic sequence beginning with a and increasing by d, i.e., a, a + d, a + 2d, a + 3d, a

poj3006 Dirichlet's Theorem on Arithmetic Progressions

Dirichlet's Theorem on Arithmetic Progressions Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 16333 Accepted: 8205 Description If a and d are relatively prime positive integers, the arithmetic sequence beginning with a and increasing

【POJ3006】Dirichlet's Theorem on Arithmetic Progressions（素数筛法）

简单的暴力筛法就可. 1 #include <iostream> 2 #include <cstring> 3 #include <cmath> 4 #include <cctype> 5 #include <cstdio> 6 #include <cmath> 7 #include <algorithm> 8 #include <numeric> 9 using namespace std; 10 11 co

判断素数——Dirichlet's Theorem on Arithmetic Progressions

洛谷P1214 [USACO1.4]等差数列 Arithmetic Progressions

P1214 [USACO1.4]等差数列 Arithmetic Progressions• o 156通过o 463提交• 题目提供者该用户不存在• 标签USACO• 难度普及+/提高提交讨论题解最新讨论• 这道题有问题• 怎么进一步优化时间效率啊 …题目描述一个等差数列是一个能表示成a, a+b, a+2b,..., a+nb (n=0,1,2,3,...)的数列.在这个问题中a是一个非负的整数,b是正整数.写一个程序来找出在双平方数集合(双平方数集合是所有能表示成p的平方 + q的平

POJ 3006 Dirichlet's Theorem on Arithmetic Progressions 素数 难度:0

POJ 3006 Dirichlet's Theorem on Arithmetic Progressions 素数 难度:0的相关文章

POJ 3006 Dirichlet's Theorem on Arithmetic Progressions 素数难度:0

POJ 3006 Dirichlet's Theorem on Arithmetic Progressions 素数难度:0的相关文章