We say that x is a perfect square if, for some integer b, x = b2. Similarly, x is a perfect cube if, for some integer b, x = b3. More generally, x is a perfect pth power if, for some integer b, x = bp. Given an integer x you are to determine the larg…

UVA 10622 - Perfect P-th Powers 题目链接 题意:求n转化为b^p最大的p值 思路:对n分解质因子,然后取全部质因子个数的gcd就是答案,可是这题有个坑啊.就是输入的能够是负数,负数的情况比較特殊,p仅仅能为奇数.这时候是要把答案不断除2除到为奇数就可以. 代码: #include <stdio.h> #include <string.h> #include <math.h> long long n; int prime[333333],…

Perfect Pth Powers Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 16383   Accepted: 3712 Description We say that x is a perfect square if, for some integer b, x = b2. Similarly, x is a perfect cube if, for some integer b, x = b3. More g…

题目链接: https://cn.vjudge.net/problem/POJ-1730 题目描述: We say that x is a perfect square if, for some integer b, x = b 2. Similarly, x is a perfect cube if, for some integer b, x = b 3. More generally, x is a perfect pth power if, for some integer b, x =…

We say that x is a perfect square if, for some integer b, x = b 2. Similarly, x is a perfect cube if, for some integer b, x = b 3. More generally, x is a perfect pth power if, for some integer b, x = b p. Given an integer x you are to determine the l…

这个有2种方法. 一种是通过枚举p的值(p的范围是从1-32),这样不会超时,再就是注意下精度用1e-8就可以了,还有要注意负数的处理…… #include<iostream> #include<stdio.h> #include<algorithm> #include<cmath> #include<iomanip> #include<string> using namespace std; int fun(double n) {…

题意: 对于32位有符号整数x,将其写成x = bp的形式,求p可能的最大值. 分析: 将x分解质因数,然后求所有指数的gcd即可. 对于负数还要再处理一下,负数求得的p必须是奇数才行. #include <cstdio> #include <cmath> ; ]; ], cnt = ; void Init() { int m = sqrt(maxn + 0.5); ; i <= m; ++i) if(!vis[i]) for(int j = i * i; j <= m…

留坑(p.343) 完全不知道哪里有问题qwq 从31向下开始枚举p,二分找存在性,或者数学函数什么的也兹辞啊 #include<cstdio> #include<cstring> #include<cstdlib> #include<algorithm> #include<iostream> using namespace std; void setIO(const string& s) { freopen((s + ".in…

https://vjudge.net/problem/UVA-10622 将n分解质因数,指数的gcd就是答案 如果n是负数,将答案除2至奇数 原理:(a*b)^p=a^p*b^p #include<cmath> #include<cstdio> #include<algorithm> #define N 65550 using namespace std; int gcd(int a,int b) { return !b ? a : gcd(b,a%b); } int…

题意:x可以表示为bp, 求这个p的最大值,比如 25=52, 64=26,  然后输入x 输出 p 就是一个质因子分解.算法.(表示数据上卡了2个小时.) 合数质因子分解模板. ]; ]; ; ;n!=;++i) { ) { num[cnt]=i; ){ind[cnt]++;n/=i;} cnt++; } )break; } ){num[cnt]=n; ind[cnt++]=;} 两种方法: 方法一:时间最坏的时间复杂度是(大概10^8*n)就是这种方法,数据卡了很久,如果数据再给狠一点肯定不…

/* 以前做的一道水题,再做精度控制又出了错///... */ 题目大意: 求最大完全平方数,一个数b(不超过int范围),n=b^p,使得给定n,p最大: 题目给你一个数n,求p : 解题思路: 不需要遍历b,只需要从31开始遍历p就好了.这个方法涉及到我以前过分逃避的精度控制问题:本题会使用函数pow 而pow的返回值是double转化成int 会有损失,比如4的double表示可以使4.00000000或者3.99999999999:而我们使用 强制类型转换会截取整数部分结果就可能是3,因…

http://poj.org/problem?id=1730 题意:给出一个n,a=b^p,求出最大p值. 思路: 首先利用唯一分解定理,把n写成若干个素数相乘的形势.接下来对于每个指数求最大公约数,该公约数就是所能到达的最大p值. 有一点要注意的是如果n为负数的话,如果当前p值为偶数,就一直除2直到p为奇数. #include<iostream> #include<algorithm> #include<string> #include<cstring>…

                                                         id=1730">Perfect Pth Powers Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 16746   Accepted: 3799 Description We say that x is a perfect square if, for some integer b, x = b2.…

We say that x is a perfect square if, for some integer b, x = b 2. Similarly, x is a perfect cube if, for some integer b, x = b 3. More generally, x is a perfect pth power if, for some integer b, x = b p. Given an integer x you are to determine the l…

A prime number p≥2 is an integer which is evenly divisible by only two integers: 1 and p. A composite integer is one which is not prime. The fundamental theorem of arithmetic says that any integer x can be expressed uniquely as a set of prime factors…

Input The first line of input consists of an integers T where 1≤T≤1000, the number of test cases. Then follow T lines, each containing four integers a, n, b, m satisfying 1≤n,m≤10e9, 0≤a<n, 0≤b<m. Also, you may assume gcd(n,m)=1.Output For each test…

Freddy practices various kinds of alternative medicine, such as homeopathy. This practice is based on the belief that successively diluting some substances in water or alcohol while shaking them thoroughly produces remedies for many diseases. This ye…

The factorial function, n! is defined thus for n a non-negative integer: 0! = 1 n! = n * (n-1)! (n > 0) We say that a divides b if there exists an integer k such that k*a = b Input The input to your program consists of several lines, each containing…

Input The first line of input contains one integer, giving the number of operations to perform. Then follow the operations, one per line, each of the form x1 y1 op x2 y2. Here, −109≤x1,y1,x2,y2<109 are integers, indicating that the operands are x1/y1…

You start with a sequence of consecutive integers. You want to group them into sets. You are given the interval, and an integer P. Initially, each number in the interval is in its own set. Then you consider each pair of integers in the interval. If t…

Given a sequence of positive integers, count all contiguous subsequences (sometimes called substrings, in contrast to subsequences, which may leave out elements) the sum of which is divisible by a given number. These subsequences may overlap. For exa…

The ministers of the cabinet were quite upset by the message from the Chief of Security stating that they would all have to change the four-digit room numbers on their offices. — It is a matter of security to change such things every now and then, to…

题目意思就是已知n的阶乘,求n. 当输入的阶乘小于10位数的时候,我们可以用long long将字符串转化成数字,直接计算. 而当输入的阶乘很大的时候,我们就可以利用位数去大概的估计n. //Asimple #include <bits/stdc++.h> using namespace std; typedef long long ll; ll n, m, num, res, ans, len;string str; void input() { cin >> str; len…

In the game of chess, the queen is a powerful piece. It can attack by moving any number of spaces in its current row, in its column or diagonally. In the eight queens puzzle, eight queens must be placed on a standard 8×8 chess board so that no queen…

这个东西先放在这吧.做过的以后会用#号标示出来 1.burnside定理,polya计数法    这个大家可以看brudildi的<组合数学>,那本书的这一章写的很详细也很容易理解.最好能完全看懂了,理解了再去做题,不要只记个公式.    *简单题:(直接用套公式就可以了)    pku2409 Let it Bead      #http://acm.pku.edu.cn/JudgeOnline/problem?id=2409    pku2154 Color   #http://acm.p…

转自:http://blog.sina.com.cn/s/blog_6635898a0100magq.html 1.burnside定理,polya计数法 这个大家可以看brudildi的<组合数学>,那本书的这一章写的很详细也很容易理解.最好能完全看懂了,理解了再去做题,不要只记个公式. *简单题:(直接用套公式就可以了) pku2409 Let it Bead      http://acm.pku.edu.cn/JudgeOnline/problem?id=2409 pku2154 Co…

发信人: fennec (fennec), 信区: Algorithm 标 题: acm 总结 by fennec 发信站: 吉林大学牡丹园站 (Wed Dec 8 16:27:55 2004) ACM总结(fennec) 其实在北京比赛完的时候,我就想写了,不过还是早了点,直到上海比赛结束,大家的心中都不是太好受.郭老师有句话:你们这样做也是对的,不成功就成仁.让我的心也能安慰了不少. 我是从大一下学期开始接触ACM的,那时候我们学校才刚刚起步,siyee,wjiang师兄可以说是我的领路人了…

各种杂题,水题,模拟,包括简单数论. 1001 A+B 1002 A+B+C 1009 Fat Cat 1010 The Angle 1011 Unix ls 1012 Decoding Task 1019 Grandpa's Other Estate 1034 Simple Arithmetics 1036 Complete the sequence! 1043 Maya Calendar 1054 Game Prediction 1057 Mileage Bank 1067 Rails 10…

 1.burnside定理,polya计数法 这个专题我单独写了个小结,大家可以简单参考一下:polya 计数法,burnside定理小结 2.置换,置换的运算 置换的概念还是比较好理解的,<组合数学>里面有讲.对于置换的幂运算大家可以参考一下潘震皓的那篇<置换群快速幂运算研究与探讨>,写的很好. *简单题:(应该理解概念就可以了) pku3270 Cow Sorting http://acm.pku.edu.cn/JudgeOnline/problem?id=3270 pku…

图论 图论解题报告索引 DFS poj1321 - 棋盘问题 poj1416 - Shredding Company poj2676 - Sudoku poj2488 - A Knight's Journey poj1724 - ROADS(邻接表+DFS) BFS poj3278 - Catch That Cow(空间BFS) poj2251 - Dungeon Master(空间BFS) poj3414 - Pots poj1915 - Knight Moves poj3126 - Prim…