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…

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: 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 and then,…

Prime Distance Time Limit: 2 Seconds      Memory Limit: 65536 KB The branch of mathematics called number theory is about properties of numbers. One of the areas that has captured the interest of number theoreticians for thousands of years is the ques…

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…

题目大意 给定两个数L和U,要求你求出在区间[L, U] 内所有素数中,相邻两个素数差值最小的两个素数C1和C2以及相邻两个素数差值最大的两个素数D1和D2,并且L-U<1,000,000 题解 由于1<=L< U<=2,147,483,647,直接筛肯定超时,但是题目说L-U<1,000,000,我们可以先筛选出sqrt(2147483647)(约等于46340)内的素数即可,然后再用这些素数把区间[L,U]内的合数筛选掉,最后就可以枚举答案了~~~ 代码: #includ…

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…

Problem Description Everybody knows any number can be combined by the prime number. Now, your task is telling me what position of the largest prime factor. The position of prime 2 is 1, prime 3 is 2, and prime 5 is 3, etc. Specially, LPF(1) = 0. Inpu…

描述 孤单的zydsg又一次孤单的度过了520,不过下一次不会再这样了.zydsg要做些改变,他想去和素数小姐姐约会. 所有的路口都被标号为了一个4位素数,zydsg现在的位置和素数小姐姐的家也是这样,如果两个路口间只差1个数字,则有一条路连通两个路口.(例如1033和1073间有一条路连接) 现在,你知道了zydsg的位置和素数小姐姐的家,问最少zydsg要走多少条路才能见到素数小姐姐.例如:如果zydsg在1033,素数小姐姐的家在8179,最少要走6条街,走法为: 1033 1733 37…