hdu 1905 Pseudoprime numbers】的更多相关文章

#include<stdio.h> #include<math.h> #define ll long long ll mod; bool Judge(int x) { ;i<sqrt(x+0.1);i++) { ) return true; } return false; } ll mult(ll q,ll n) { ll ret=q; ll ans=; ) { ) { ans*=ret; ans%=mod; } ret=(ret*ret)%mod; n>>=;…
Pseudoprime numbers Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 11336   Accepted: 4891 Description Fermat's theorem states that for any prime number p and for any integer a > 1, ap = a (mod p). That is, if we raise a to the pth power…
Problem Description Fermat's theorem states that for any prime number p and for any integer a > 1, a^p == a (mod p). That is, if we raise a to the pth power and divide by p, the remainder is a. Some (but not very many) non-prime values of p, known as…
题目连接 http://poj.org/problem?id=3641 Pseudoprime numbers Description Fermat's theorem states that for any prime number p and for any integer a > 1, ap = a (mod p). That is, if we raise a to the pth power and divide by p, the remainder is a. Some (but…
POJ3641 Pseudoprime numbers p是Pseudoprime numbers的条件: p是合数,(p^a)%p=a;所以首先要进行素数判断,再快速幂. 此题是大白P122 Carmichael Number 的简化版 /* * Created: 2016年03月30日 22时32分15秒 星期三 * Author: Akrusher * */ #include <cstdio> #include <cstdlib> #include <cstring&g…
原题直通车: HDU  4722  Good Numbers 题意: 求区间[a,b]中各位数和mod 10==0的个数. 代码: #include<iostream> #include<cstdio> #include<cstring> #include<cmath> using namespace std; int f[20]; long long work(long long x){ long long ret=0, u=x; int t=0, s=0…
HDU 3117 Fibonacci Numbers(斐波那契前后四位,打表+取对+矩阵高速幂) ACM 题目地址:HDU 3117 Fibonacci Numbers 题意:  求第n个斐波那契数的前四位和后四位.  不足8位直接输出. 分析:  前四位有另外一题HDU 1568,用取对的方法来做的.  后四位能够用矩阵高速幂,MOD设成10000即可了. 代码: /* * Author: illuz <iilluzen[at]gmail.com> * Blog: http://blog.c…
HDOJ(HDU).1058 Humble Numbers (DP) 点我挑战题目 题意分析 水 代码总览 /* Title:HDOJ.1058 Author:pengwill Date:2017-2-15 */ #include <iostream> #include <cstdio> #include <algorithm> #include <cstring> #define nmax 6000 #define min(a,b) (a<b?a:b…
Pseudoprime numbers Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 10903   Accepted: 4710 Description Fermat's theorem states that for any prime number p and for any integer a > 1, ap = a (mod p). That is, if we raise a to the pth power…
Pseudoprime numbers Descriptions 费马定理指出,对于任意的素数 p 和任意的整数 a > 1,满足 ap = a (mod p) .也就是说,a的 p 次幂除以 p 的余数等于 a .p 的某些 (但不是很多) 非素数的值,被称之为以 a 为底的伪素数,对于某个 a 具有该特性.并且,某些 Carmichael 数,对于全部的 a 来说,是以 a为底的伪素数. 给定 2 < p ≤ 1000000000 且 1 < a < p ,判断 p 是否为以 …