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…

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 <= ; ++…

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 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…

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…

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>…