埃拉托斯特尼筛法(sieve of Eratosthenes ) 是古希腊数学家埃拉托斯特尼发明的计算素数的方法.对于求解不大于n的所有素数,我们先找出sqrt(n)内的所有素数p1到pk,其中k = sqrt(n),依次剔除Pi的倍数,剩下的所有数都是素数. 具体操作如上述 图片所示. C++实现 #include<iostream> #include<vector> using namespace std; int main() { int n; cin >> n;…

题目: http://poj.org/problem?id=3126 困得不行了,没想到敲完一遍直接就A了,16ms,debug环节都没进行.人品啊. #include <stdio.h> #include <string.h> #include <queue> using namespace std; ]; ]; int s, t; void prime_init() { memset(prime, , sizeof(prime)); prime[] = ; ; i…

Prime Path(POJ - 3126) 题目链接 算法 BFS+筛素数打表 1.题目主要就是给定你两个四位数的质数a,b,让你计算从a变到b共最小需要多少步.要求每次只能变1位,并且变1位后仍然为质数. 2.四位数的范围是1000~9999,之间共有1000多个质数.由于已经知道位数为4位,所以可以通过BFS来寻找最小步数.每次需要分别变换个位.十位.百位.千位,并且把符合要求的数放到队列中,同时需标记这个数已经遍历过一次,避免重复遍历,直到找到目标数. C++代码 #include<io…

Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 9519   Accepted: 5458 Description The ministers of the cabinet were quite upset by the message from the Chief of Security stating that they would all have to change the four-digit room numb…

Prime Path The ministers of the cabinet were quite upset by the message from the Chief of Security stating that they would all have to change the four-digit room numbers on their offices. — It is a matter of security to change such things every now a…

Language: Default Prime Path Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 11703   Accepted: 6640 Description The ministers of the cabinet were quite upset by the message from the Chief of Security stating that they would all have to c…

Prime Path Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 27132   Accepted: 14861 Description The ministers of the cabinet were quite upset by the message from the Chief of Security stating that they would all have to change the four-di…

Prime Path Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 15325   Accepted: 8634 Description The ministers of the cabinet were quite upset by the message from the Chief of Security stating that they would all have to change the four-dig…

/* * 二次筛素数 * POJ268----Prime Distance(数论,素数筛) */ #include<cstdio> #include<vector> using namespace std; const int maxn = 1000005; typedef long long LL; bool is_prime_small[maxn]; bool is_prime[maxn]; vector <int> res; int main() { LL l,u…

题目链接:http://poj.org/problem?id=3126 题意: 给定两个四位素数 $a,b$,要求把 $a$ 变换到 $b$.变换的过程每次只能改动一个数,要保证每次变换出来的数都是一个没有前导零的四位素数. 要求每步得到的素数都不能重复,求从 $a$ 到 $b$ 最少需要变换多少步:如果无法达到则输出Impossible. 题解: 在BFS之前先用线性筛筛出 $10000$ 以内的素数,方便后面判断是否为素数. 剩下的就是从 $a$ 为起点,入队并标记已经出现过.每次队列非空就…