题意: 就是哥德巴赫猜想...任意一个偶数 都可以分解成两个(就是一对啦)质数的加和 输入一个偶数求有几对.. 解析: 首先! 素数打表..因为 质数 + 质数 = 偶数 所以 偶数 - 质数 = 质数 ... 我真是蠢啊 还有  vis要用bool类型的!!!!  int会直接爆 代码如下: #include <iostream> #include <cstdio> #include <sstream> #include <cstring> #includ…

Goldbach`s Conjecture(LightOJ - 1259)[简单数论][筛法] 标签: 入门讲座题解 数论 题目描述 Goldbach's conjecture is one of the oldest unsolved problems in number theory and in all of mathematics. It states: Every even integer, greater than 2, can be expressed as the sum of…

链接: https://vjudge.net/problem/LightOJ-1259 题意: Goldbach's conjecture is one of the oldest unsolved problems in number theory and in all of mathematics. It states: Every even integer, greater than 2, can be expressed as the sum of two primes [1]. Now…

[LightOJ1259]Goldbach`s Conjecture(数论) 题面 Vjudge T组询问,每组询问是一个偶数n 验证哥德巴赫猜想 回答n=a+b 且a,b(a<=b)是质数的方案个数 题解 筛出质数后直接暴力判断就行了 质数密度没有那么大,记得节约空间 #include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<cmath>…

Pairs Forming LCM (LightOJ - 1236)[简单数论][质因数分解][算术基本定理](未完成) 标签: 入门讲座题解 数论 题目描述 Find the result of the following code: long long pairsFormLCM( int n ) { long long res = 0; for( int i = 1; i <= n; i++ ) for( int j = i; j <= n; j++ ) if( lcm(i, j) ==…

Help Hanzo (LightOJ - 1197) [简单数论][筛区间质数] 标签: 入门讲座题解 数论 题目描述 Amakusa, the evil spiritual leader has captured the beautiful princess Nakururu. The reason behind this is he had a little problem with Hanzo Hattori, the best ninja and the love of Nakurur…

Aladdin and the Flying Carpet (LightOJ - 1341)[简单数论][算术基本定理][分解质因数](未完成) 标签:入门讲座题解 数论 题目描述 It's said that Aladdin had to solve seven mysteries before getting the Magical Lamp which summons a powerful Genie. Here we are concerned about the first myste…

Sigma Function (LightOJ - 1336)[简单数论][算术基本定理][思维] 标签: 入门讲座题解 数论 题目描述 Sigma function is an interesting function in Number Theory. It is denoted by the Greek letter Sigma (σ). This function actually denotes the sum of all divisors of a number. For exam…

In 1742, Christian Goldbach, a German amateur mathematician, sent a letter to Leonhard Euler in which he made the following conjecture: Every even number greater than 4 can be written as the sum of two odd prime numbers. For example: 8 = 3 + 5. Both…

Goldbach's conjecture is one of the oldest unsolved problems in number theory and in all of mathematics. It states: Every even integer, greater than 2, can be expressed as the sum of two primes [1]. Now your task is to check whether this conjecture h…

一.Description In 1742, Christian Goldbach, a German amateur mathematician, sent a letter to Leonhard Euler in which he made the following conjecture: Every even number greater than 4 can be written as the sum of two odd prime numbers. For example: 8…

题目链接:https://vjudge.net/problem/LightOJ-1259 1259 - Goldbach`s Conjecture    PDF (English) Statistics Forum Time Limit: 2 second(s) Memory Limit: 32 MB Goldbach's conjecture is one of the oldest unsolved problems in number theory and in all of mathem…

                                                                                1259 - Goldbach`s Conjecture 分拆素数和,本来很水的题,但硬生生跪了18+.这题卡内存实在太凶残了,MLT了10+,TLE了10+. 题意:在1e7内验证哥德巴赫猜想,求一个数拆分成两个不同的素数和有多少中方法. 思路:打表.就是打表. 但为什么会MLE呢,因为1e7的数组实在太大了,于是换了好几种写法优化,但…

Goldbach's Conjecture Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 48161   Accepted: 18300 Description In 1742, Christian Goldbach, a German amateur mathematician, sent a letter to Leonhard Euler in which he made the following conject…

[题目描述] 原题来自:Ulm Local,题面详见:POJ 2262 哥德巴赫猜想:任何大于 44 的偶数都可以拆成两个奇素数之和. 比如: 8=3+5 20=3+17=7+13 42=5+37=11+31=13+29=19+23 你的任务是:验证小于 106 的数满足哥德巴赫猜想. [输入] 多组数据,每组数据一个 n. 读入以 0 结束. [输出] 对于每组数据,输出形如 n=a+b,其中 a,b 是奇素数.若有多组满足条件的 a,b,输出 b−a 最大的一组. 若无解,输出 Goldba…

 Goldbach's Conjecture Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Submit Status Description In 1742, Christian Goldbach, a German amateur mathematician, sent a letter to Leonhard Euler in which he made the following…

1.Link: http://poj.org/problem?id=2262 http://bailian.openjudge.cn/practice/2262 2.Content: Goldbach's Conjecture Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 37791   Accepted: 14536 Description In 1742, Christian Goldbach, a German a…

http://poj.org/problem?id=2262 Goldbach's Conjecture Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 34323   Accepted: 13169 Description In 1742, Christian Goldbach, a German amateur mathematician, sent a letter to Leonhard Euler in whic…

Problem Description Goldbach's Conjecture: For any even number n greater than or equal to 4, there exists at least one pair of prime numbers p1 and p2 such that n = p1 + p2. This conjecture has not been proved nor refused yet. No one is sure whether…

Goldbach's Conjecture Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 5277    Accepted Submission(s): 2022点我 Problem Description Goldbach's Conjecture: For any even number n greater than or equa…

题目大意:输入一个整数n,输出使2^x mod n = 1成立的最小值K 解题思路:简单数论 1)n可能不能为偶数.因为偶数可不可能模上偶数以后==1. 2)n肯定不可能为1 .因为任何数模上1 == 0: 3)所以n肯定是除1外的奇数 代码如下: #include <iostream> using namespace std; int main(){ int n; while(scanf("%d",&n)!=EOF){ if(n == 1 || n % 2 ==…

题目大意:给一个偶数,判断是否是两个素数的和. 先用sieve方法生成一个素数表,然后再进行判断即可. #include <cstdio> #include <vector> #include <bitset> using namespace std; typedef vector<int> vi; typedef long long ll; #define MAXN 1000000 vi primes; bitset<MAXN+> bs; vo…

[整除] 若a被b整除,即a是b的倍数,那么记作b|a("|"是整除符号),读作"b整除a"或"a能被b整除".b叫做a的约数(或因数),a叫做b的倍数. [质因数分解] 把一个正整数数分解成几个质数的幂相乘的形式叫做质因数分解. e.g. 10=2*5 16=24 18=2*32 [唯一分解定理] 唯一分解定理(算术基本定理)可表述为:任何一个大于1的自然数 N,如果N不为质数,那么N可以唯一分解成有限个质数的乘积: N=P1a1*P2a2*P…

传送门 简单数论暴力题. 题目简述:要求求出所有满足x2≡1mod&ThinSpace;&ThinSpace;nx^2\equiv1 \mod nx2≡1modn且0≤x<n0\le x<n0≤x<n的xxx 考虑到使用平方差公式变形. (x−1)(x+1)≡0mod&ThinSpace;&ThinSpace;n(x-1)(x+1)\equiv0 \mod n(x−1)(x+1)≡0modn 即(x−1)(x+1)=kn(x-1)(x+1)=kn(x−1)…

题目链接: https://cn.vjudge.net/problem/POJ-2262 题目描述: In 1742, Christian Goldbach, a German amateur mathematician, sent a letter to Leonhard Euler in which he made the following conjecture: Every even number greater than 4 can be written as the sum of t…

题目链接:http://ac.jobdu.com/problem.php?pid=1440 详解链接:https://github.com/zpfbuaa/JobduInCPlusPlus 参考代码: // // 1440 Goldbach's Conjecture.cpp // Jobdu // // Created by PengFei_Zheng on 12/04/2017. // Copyright © 2017 PengFei_Zheng. All rights reserved. /…

Goldbach`s Conjecture Goldbach's conjecture is one of the oldest unsolved problems in number theory and in all of mathematics. It states: Every even integer, greater than 2, can be expressed as the sum of two primes [1]. Now your task is to check whe…

poj2262 Goldbach's Conjecture 用欧拉筛把素数筛出来,再枚举一下. #include<iostream> #include<cstdio> #include<cstring> #define re register using namespace std; #define N 1000001 int n,v[N],pri[N],cct; bool is[N]; int main(){ ;i<N;++i){ ; ;j<=cct;++…

解法一:数论筛法+前缀和 //其实题目中f[n]的值可理解为存在多少个整数对使a*b<=n #include<cstdio> #define N 1007 #define maxn 1000005 using namespace std; int f[maxn]; void Procede(int n)//预处理 { ;i<maxn;i++){ for(int j=i;j<maxn;j+=i){ f[j]++; } } //此时f[n]理解为表示n的约数的个数,亦可以理解为表…

题目描述 Goldbach's Conjecture: For any even number n greater than or equal to 4, there exists at least one pair of prime numbers p1 and p2 such that n = p1 + p2. This conjecture has not been proved nor refused yet. No one is sure whether this conjecture…