Problem Description Eddy's interest is very extensive, recently he is interested in prime number. Eddy discover the all number owned can be divided into the multiply of prime number, but he can't write program, so Eddy has to ask intelligent you to h…

Problem Description Eddy's interest is very extensive, recently he is interested in prime number. Eddy discover the all number owned can be divided into the multiply of prime number, but he can't write program, so Eddy has to ask intelligent you to h…

上一篇已经讲了,但是转载别人的很乱,这里自己根据blog里面写的思路,重新写过了一个程序 #include <iostream> #include <malloc.h> #include <math.h> #define N 65535 int primes[N]; int SimpleDivsion()//生成1-65535之间所有的质数 { ; int i,j; //int *primes = (int *)malloc(sizeof(int)*n); primes…

Eddy's research I Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 5793    Accepted Submission(s): 3459 Problem Description Eddy's interest is very extensive, recently he is interested in prime n…

http://acm.hdu.edu.cn/showproblem.php?pid=1164 题意很简单,只是写代码的时候需要注意几个问题 一.筛选素数的时候记得用埃式筛选法,要是直接找可能会WA. int prime(int n) { int flag = 0; for(int i=2;i<=sqrt(n);i++) { if(n%i==0){ flag = 1;break; } } if(flag==1) return 0; else return 1; } void solve() { f…

题目链接 题意 : 给你一个数,让你用它的素数质因子表示出来. 思路 : 先打一下表,因为会有重复的质因子,所以从大到小开始找,并且找到一个之后不能就接着往下找,要再找一遍这个数. #include <stdio.h> #include <string.h> #include <math.h> #include <iostream> using namespace std ; ] ; void sett() { int i,j; ; i <= ; ++…

链接:传送门 题意:给出一个整数 n ,输出整数 n 的分解成若干个素因子的方案 思路:经典的整数分解题目,这里采用试除法 和 用筛法改造后的试除法 对正整数 n 进行分解 方法一:试除法对正整数 n 进行分解 /************************************************************************* > File Name: hdu1164.cpp > Author: WArobot > Blog: http://www.cnb…

思路:将输入的这个数分成n个素数的相乘的结果,用一个数组存储起来.之后再输出就能够了 Eddy's research I Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 6633    Accepted Submission(s): 3971 Problem Description Eddy's interest is very ext…

Eddy's research I Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 7389    Accepted Submission(s): 4472 Problem Description Eddy's interest is very extensive, recently he is interested in prime n…

- Eddy's research II Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Description As is known, Ackermann function plays an important role in the sphere of theoretical computer science. However, in the other h…

并查集+最小生成树 Eddy's picture Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 7669    Accepted Submission(s): 3882 Problem Description Eddy begins to like painting pictures recently ,he is sure of hi…

Problem Description Eddy是个ACMer,他不仅喜欢做ACM题,而且对于Ranklist中每个人的ac数量也有一定的研究,他在无聊时经常在纸上把Ranklist上每个人的ac题目的数量摘录下来,然后从中选择一部分人(或者全部)按照ac的数量分成两组进行比较,他想使第一组中的最小ac数大于第二组中的最大ac数,但是这样的情况会有很多,聪明的你知道这样的情况有多少种吗? 特别说明:为了问题的简化,我们这里假设摘录下的人数为n人,而且每个人ac的数量不会相等,最后结果在64位整数…

Problem Description The digital root of a positive integer is found by summing the digits of the integer. If the resulting value is a single digit then that digit is the digital root. If the resulting value contains two or more digits, those digits a…

题意:给定一个表达式,然后让你求表达式的值. 析:多写几个就会发现规律. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring>…

题意:已知,求A(m, n). 分析:根据样例模拟一下过程就可以找出递推关系. #include<cstdio> #include<cstring> #include<cstdlib> #include<cctype> #include<cmath> #include<iostream> #include<sstream> #include<iterator> #include<algorithm>…

原文链接:http://www.cnblogs.com/DrunBee/archive/2012/09/05/2672546.html 题意:给你一个正整数N,确定在1到N之间有多少个可以表示成M^K(K>1)的数. 我们可以由n^(1/p),知道指数为p的有多少个数. 通过观察,可以发现若一个数可以表示成x^(k*t),则可以表示成(x^k)^t.因此指数必然为素数. 枚举素数便可以得到指数为p的个数,但是可能出现重复,例如:x^3=y^5,其中x=t^5,y=t^3. 运用容斥原理,设a[i…

转载请注明出处:http://blog.csdn.net/a1dark 分析:找规律 #include<stdio.h> int main(){ int m,n; while(scanf("%d%d",&m,&n)!=EOF){ if(m==1)printf("%d\n",n+2); else if(m==2)printf("%d\n",2*n+3); else{ int s=5; for(int i=1;i<=…

我在网上看了一些大牛的题解,有些知识点不是太清楚, 因此再次整理了一下. 转载链接: http://blog.csdn.net/iamskying/article/details/4738838 http://www.2cto.com/kf/201405/297531.html 题目描述:求n^n次的digital root(数根),例如root(67)=6+7=root(13)=1+3=4; 一类解法: 求解思路:现在分析一个问题,假设将十位数为a,个位数为b的一个整数表示为ab,则推导得ab…

题目:http://acm.hdu.edu.cn/showproblem.php?pid=1165 果断不擅长找规律啊,做这种题静不下心来. Ackermann function can be defined recursively as follows: 递推如上图, 0<m<=3,0<=n<=1000000,,当m==3时,n>=0&&n<=24. 首先发现a(0,i)=i+1; 另外n==0时,a(1,0)=a(0,1)=2; 当m==1,n>…

Eddy's mistakes[HDU1161] Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6014    Accepted Submission(s): 3405 Problem DescriptionEddy usually writes articles ,but he likes mixing the English let…

Eddy's research II Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2706    Accepted Submission(s): 985 Problem Description As is known, Ackermann function plays an important role in the sphere o…

Eddy's research I Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 5050    Accepted Submission(s): 3027 Problem Description Eddy's interest is very extensive, recently he is interested in prime n…

Eddy's research I Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 7537 Accepted Submission(s): 4579 Problem Description Eddy's interest is very extensive, recently he is interested in prime number…

Eddy's research I Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 6664    Accepted Submission(s): 3997 Problem Description Eddy's interest is very extensive, recently he is interested in prime…

这个分类怎么觉得这么水呢.. 这个分类做到尾的模板集: //gcd int gcd(int a,int b){return b? gcd(b, a % b) : a;} //埃氏筛法 O(nlogn) int prime[MAX_N]; bool is_prime[MAX_N];//i是不是素数 int sieve() { ; ; i <= n; i++) is_prime[i] = true; is_prime[] = is_prime[] = false; ; i <= n; i++) {…

Eddy's research I Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 7402    Accepted Submission(s): 4479 Problem Description Eddy's interest is very extensive, recently he is interested in prime…

题目链接: acm.hdu.edu.cn/showproblem.php?pid=1165 Eddy's research II Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 5656    Accepted Submission(s): 2045 Problem Description As is known, Ackermann f…

Problem Description Eddy usually writes articles ,but he likes mixing the English letter uses, for example "computer science" is written frequently "coMpUtEr scIeNce" by him, this mistakes lets Eddy's English teacher be extremely disco…

// Eddy 继续 Problem Description As is known, Ackermann function plays an important role in the sphere of theoretical computer science. However, in the other hand, the dramatic fast increasing pace of the function caused the value of Ackermann function…

Problem Description Eddy usually writes  articles ,but he likes mixing the English letter uses, for example "computer science" is written frequently "coMpUtEr scIeNce" by him, this mistakes lets Eddy's English teacher be extremely disc…