http://poj.org/problem?id=1811 #include <cstdio> #include <cstring> #include <algorithm> #include <ctime> using namespace std; typedef __int64 LL; ; LL minf, n; LL random(LL n){ return (double)rand() / RAND_MAX * n + 0.5; } LL mult…

http://blog.csdn.net/shiyuankongbu/article/details/9202373 发现自己原来的那份模板是有问题的,而且竟然找不出是哪里的问题,所以就用了上面的链接上的一份代码,下面只是寄存一下这份代码,以后打印出来当模板好了. #pragma warning(disable:4996) #include <iostream> #include <cstring> #include <algorithm> #include <c…

问题描述:素性测试兼质因子分解 解题关键:pollard-rho质因数分解,在RSA的破译中也起到了很大的作用 期望复杂度:$O({n^{\frac{1}{4}}})$ #include<cstdio> #include<cstring> #include<algorithm> #include<cstdlib> #include<iostream> #include<cmath> #include<vector> #de…

Description Given a big integer number, you are required to find out whether it's a prime number. Input The first line contains the number of test cases T (1 <= T <= 20 ), then the following T lines each contains an integer number N (2 <= N <…

[题目大意] 若n是素数,输出“Prime”,否则输出n的最小素因子,(n<=2^54) [题解] 和bzoj3667差不多,知识这道题没那么坑. 直接上Pollord_Rho和Rabin_Miller就行了. /************* POJ 1811 by chty 2016.11.7 *************/ #include<iostream> #include<cstdio> #include<cstdlib> #include<cstri…

Description: Count the number of prime numbers less than a non-negative number, n click to show more hints. Credits:Special thanks to @mithmatt for adding this problem and creating all test cases. 求n以内的所有素数,以前看过的一道题目,通过将所有非素数标记出来,再找出素数,代码如下: public i…

Peter wants to generate some prime numbers for his cryptosystem. Help him! Your task is to generate all prime numbers between two given numbers! Input The input begins with the number t of test cases in a single line (t<=10). In each of the next t li…

Sum of Consecutive Prime Numbers Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 20050   Accepted: 10989 Description Some positive integers can be represented by a sum of one or more consecutive prime numbers. How many such representatio…

传送门 Description A ring is composed of n (even number) circles as shown in diagram. Put natural numbers 1, 2, . . . , n into each circle separately, and the sum of numbers in two adjacent circles should be a prime. Note: the number of first circle sho…

题意为给出两个四位素数A.B,每次只能对A的某一位数字进行修改,使它成为另一个四位的素数,问最少经过多少操作,能使A变到B.可以直接进行BFS搜索 #include<bits/stdc++.h> using namespace std; bool isPrime(int n){//素数判断 || n == ) return true; else{ ; ; i < k; i++){ ) return false; } return true; } } ]; ]; void getPrime…