题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5108 题目意思:给出一个数正整数 N,N <= 1e9,现在需要找出一个最少的正整数 M,使得 N/M 是素数.如果找不到就输出0. 一开始有想过将所有 <= 1e9 的素数求出来的,不过绝对超时就放弃了:然后就开始从题目中挖掘简便的处理方法.受到求素数的方法启发,枚举的因子 i 如果在 i * i <= N 之内都没有找到符合条件的素数,那么那些 > N 的因子就更不可能了.于是时间…
Alexandra and Prime Numbers Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1847 Accepted Submission(s): 629 Problem Description Alexandra has a little brother. He is new to programming. One…
题意: 给一个正整数N,找最小的M,使得N可以整除M,且N/M是质数. 数据范围: There are multiple test cases (no more than 1,000). Each case contains only one positive integer N.N≤1,000,000,000.Number of cases with N>1,000,000 is no more than 100. 思路: N=M*prime 故必有M或prime小于等于sqrt(N)…
Saving HDU Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 7194 Accepted Submission(s): 3345 Problem Description 话说上回讲到海东集团面临内外交困,公司的元老也只剩下XHD夫妇二人了.显然,作为多年拼搏的商人,XHD不会坐以待毙的. 一天,当他正在苦思冥想解困良策的…
http://acm.hdu.edu.cn/showproblem.php?pid=3037 Lucas定理模板. 现在才写,noip滚粗前兆QAQ #include<cstdio> #include<cstring> #include<algorithm> using namespace std; typedef long long ll; int jc[100003]; int p; int ipow(int x, int b) { ll t = 1, w = x;…
Special equations Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u Submit Status Practice HDU 4569 Description Let f(x) = a nx n +...+ a 1x +a 0, in which a i (0 <= i <= n) are all known integers. We call f(x) 0 (mod…
The kth great number Time Limit:1000MS Memory Limit:65768KB 64bit IO Format:%I64d & %I64u Submit Status Practice HDU 4006 Description Xiao Ming and Xiao Bao are playing a simple Numbers game. In a round Xiao Ming can choose to write down a nu…
How many integers can you find Time Limit:5000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u Submit Status Practice HDU 1796 Description Now you get a number N, and a M-integers set, you should find out how many integers which are sm…
hdu5901题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5901 code vs 3223题目链接:http://codevs.cn/problem/3223/ 思路:主要是用了一个Meisell-Lehmer算法模板,复杂度O(n^(2/3)).讲道理,我不是很懂(瞎说什么大实话....),下面输出请自己改 #include<bits/stdc++.h> using namespace std; typedef long long LL;…