题目 其实这道题不是很难,但是我刚开始拿到这道题的时候不知道怎么做, 因为这个式子我就不知道是干什么的: 65|f(x) 百度解释(若a/b=x...0  称a能被b整除,b能整除a,即b|a,读作"b整除a"或"a能被b整除".a叫做b的倍数,b叫做a的约数(或因数).) 即:f(x)能够被65整除. 即题目大意是: 方程f(x)=5*x^13+13*x^5+k*a*x:输入任意一个数k,是否存在一个数a,对任意x都能使得f(x)能被65整除 解题思路: 当x=1…

Ignatius's puzzle Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 4935    Accepted Submission(s): 3359 Problem Description Ignatius is poor at math,he falls across a puzzle problem,so he has no…

Ignatius's puzzle Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 5035    Accepted Submission(s): 3426 Problem Description Ignatius is poor at math,he falls across a puzzle problem,so he has no…

Ignatius's puzzle Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 9934    Accepted Submission(s): 6959 Problem Description Ignatius is poor at math,he falls across a puzzle problem,so he has no…

Ignatius's puzzle Problem Description Ignatius is poor at math,he falls across a puzzle problem,so he has no choice but to appeal to Eddy. this problem describes that:f(x)=5x13+13*x5+ka*x,input a nonegative integer k(k<10000),to find the minimal none…

Problem Description Ignatius is poor at math,he falls across a puzzle problem,so he has no choice but to appeal to Eddy. this problem describes that:f(x)=5*x^13+13*x^5+k*a*x,input a nonegative integer k(k<10000),to find the minimal nonegative integer…

题目大意: 给定k,找到一个满足的a使任意的x都满足 f(x)=5*x^13+13*x^5+k*a*x 被65整除 推证: f(x) = (5*x^12 + 13 * x^4 + ak) * x 因为x可以任意取 那么不能总是满足 65|x 那么必须是 65 | (5*x^12 + 13 * x^4 + ak) 那么就是说 x^12 / 13 + x^4 / 5 + ak / 65 正好是一个整数 假设能找到满足的a , 那么将 ak / 65 分进x^12 / 13 + x^4 / 5中得到…

以下引用自http://acm.hdu.edu.cn/discuss/problem/post/reply.php?postid=8466&messageid=2&deep=1 题意以及简单证明Posted by hncu1106101-王志盼 at 2012-10-19 16:35:42 on Problem 1098 有一个函数: f(x)=5*x^13+13*x^5+k*a*x给定一个非负的 k 值 求最小的非负的 a 值 使得对任意的整数x都能使 f(x) 被 65 整除. 每输入…

http://acm.hdu.edu.cn/showproblem.php?pid=1098 题意 :输入一个K,让你找一个a,使得f(x)=5*x^13+13*x^5+k*a*x这个f(x)%65等于0. 思路: 这个题我也不是很会,看了网上的思路才做的. http://www.cnblogs.com/g0feng/archive/2012/08/23/2652996.html //HDU 1098 #include <iostream> #include <stdio.h> u…

数学归纳法,得证只需求得使18+ka被64整除的a.且a不超过65. #include <stdio.h> int main() { int i, j, k; while (scanf("%d", &k) != EOF) { j = ; ; i<; i++) { +k*i) % == ) { j = ; break; } } if (j) printf("%d\n", i); else printf("no\n"); }…

链接 [http://acm.hdu.edu.cn/showproblem.php?pid=1098] 分析: 数学归纳法 f(1) = 18 + ka; 假设f(x) = 5x^13+13x^5+kax 能被65整除 f(x+1) = 5(x+1)^13+13(x+1)^5+ka(x+1) 根据二项式定理展开 (a+b)^n = C(n,0)a^nb^0 + C(n,1)a^(n-1)b^1 + C(n,2)a^(n-2)b^2 + ... + C(n,n)a^0b^n f(x+1) = 5…

http://acm.hdu.edu.cn/showproblem.php?pid=1098 其实一开始猜测只要验证x=1的时候就行了,但是不知道怎么证明. 题解表示用数学归纳法,假设f(x)成立,证明f(x+1)成立需要什么条件. 代入之后发现有很多二项式系数,导致他们都是65的倍数,剩下的恰好就是 f(x) 和 18+ka . 那么只需要找到最小的a使得 18+ka是65的倍数. 题解说,毕竟65毕竟小,可以枚举a.因为a+65与a的对65的余数是一样的,所以只要枚举0到64就可以了. 我的…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1098 题目中文是这样的: 伊格内修斯在数学上很差,他遇到了一个难题,所以他别无选择,只能上诉埃迪. 这个问题描述:f(x)= 5 * x ^ 13 + 13 * x ^ 5 + k * a * x,输入一个非正整数k(k <10000),找到最小非负整数a, 使得任意的整数x,65 | f(x)如果不存在那个a,就打印“否”. 输入 输入包含多个测试用例. 每个测试用例由非负整数k组成,样例输入中有…

杭电ACM分类: 1001 整数求和 水题1002 C语言实验题——两个数比较 水题1003 1.2.3.4.5... 简单题1004 渊子赛马 排序+贪心的方法归并1005 Hero In Maze 广度搜索1006 Redraiment猜想 数论:容斥定理1007 童年生活二三事 递推题1008 University 简单hash1009 目标柏林 简单模拟题1010 Rails 模拟题(堆栈)1011 Box of Bricks 简单题1012 IMMEDIATE DECODABILITY…

Ignatius's puzzle Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 9859    Accepted Submission(s): 6898 Problem Description Ignatius is poor at math,he falls across a puzzle problem,so he has no…

转载:from http://blog.csdn.net/qq_28236309/article/details/47818349 基础题:1000.1001.1004.1005.1008.1012.1013.1014.1017.1019.1021.1028.1029. 1032.1037.1040.1048.1056.1058.1061.1070.1076.1089.1090.1091.1092.1093. 1094.1095.1096.1097.1098.1106.1108.1157.116…

分析:裴蜀定理,a,b互质的充要条件是存在整数x,y使ax+by=1.存在整数x,y,使得ax+by=c.那么c就是a,b的公约数. 如果存在数a ,由于对随意x方程都成立.则有当x=1时f(x)=18+ka;有由于f(x)能被65整除,所以f(x)=n*65.即18+ka=n*65有整数解则说明如果成立. ax+by = c的方程有整数解的一个充要条件是:c%gcd(a, b) == 0.然后枚举直到(65*n-18)%k == 0. #include<iostream> using nam…

这个话题.简单的数学. 对于函数,f(x)=5*x^13+13*x^5+k*a*x,输入k,对于休闲x,一个数字的存在a,使f(x)是65可分. 对于休闲x. 因此,当x = 1时间,f(x) = 18 + a* k.满足被65整除. 也就是(18 + a * k)% 65 = 0. 所以,一切都非常easy了. 以下的是AC的代码: #include <iostream> using namespace std; int main() { int k; while(cin >>…

A hard puzzle Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 23766    Accepted Submission(s): 8390 Problem Description lcy gives a hard puzzle to feng5166,lwg,JGShining and Ignatius: gave a and…

Problem Description lcy gives a hard puzzle to feng5166,lwg,JGShining and Ignatius: gave a and b,how to know the a^b.everybody objects to this BT problem,so lcy makes the problem easier than begin.this puzzle describes that: gave a and b,how to know…

Problem Description lcy gives a hard puzzle to feng5166,lwg,JGShining and Ignatius: gave a and b,how to know the a^b.everybody objects to this BT problem,so lcy makes the problem easier than begin.this puzzle describes that: gave a and b,how to know…

A hard puzzle Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 29415    Accepted Submission(s): 10581 Problem Description lcy gives a hard puzzle to feng5166,lwg,JGShining and Ignatius: gave a a…

Problem Description lcy gives a hard puzzle to feng5166,lwg,JGShining and Ignatius: gave a and b,how to know the a^b.everybody objects to this BT problem,so lcy makes the problem easier than begin. this puzzle describes that: gave a and b,how to know…

Problem Description lcy gives a hard puzzle to feng5166,lwg,JGShining and Ignatius: gave a and b,how to know the a^b.everybody objects to this BT problem,so lcy makes the problem easier than begin.this puzzle describes that: gave a and b,how to know…

有关数论方面的题要仔细阅读,分析公式. Problem Description Ignatius is poor at math,he falls across a puzzle problem,so he has no choice but to appeal to Eddy. this problem describes that:f(x)=5*x^13+13*x^5+k*a*x,input a nonegative integer k(k<10000),to find the minima…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1097 分析:简单题,快速幂取模, 由于只要求输出最后一位,所以开始就可以直接mod10. /*A hard puzzle Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 33036 Accepted Submission(s): 11821 Pr…

A hard puzzle Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 51690 Accepted Submission(s): 18916 Problem Description lcy gives a hard puzzle to feng5166,lwg,JGShining and Ignatius: gave a and b,ho…

Problem Description lcy gives a hard puzzle to feng5166,lwg,JGShining and Ignatius: gave a and b,how to know the a^b.everybody objects to this BT problem,so lcy makes the problem easier than begin.this puzzle describes that: gave a and b,how to know…

Puzzle 面向服务/切面AOP开发框架 For .Net AOP主要实现的目的是针对业务处理过程中的切面进行提取,它所面对的是处理过程中的某个步骤或阶段,以获得逻辑过程中各部分之间低耦合性的隔离效果. 日常的产品开发中最常见的就是数据保存的功能.举例来说,现在有个用户信息数据保存的功能,我们希望在数据保存前对数据进行校验,其中现在能想到的校验就包含数据完整性校验和相同数据是否存在校验.按照传统的OOP(面向对象程序设计),我们需要定义一个IUserDataSaveService(用户数据保存…

题目 Source http://acm.hdu.edu.cn/showproblem.php?pid=5456 Description As an exciting puzzle game for kids and girlfriends, the Matches Puzzle Game asks the player to find the number of possible equations A−B=C with exactly n (5≤n≤500) matches (or stic…