Do you think that magic is simple? That some hand-waving and muttering incomprehensible blubber is enough to conjure wonderful gardens or a fireball to burn your enemies to ashes? The reality is a little more complicated than that. To master skills,…

2066. Simple Expression Time limit: 1.0 secondMemory limit: 64 MB You probably know that Alex is a very serious mathematician and he likes to solve serious problems. This is another problem from Alex. You are given three nonnegative integers a, b, c.…

C. Maximal GCD time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output You are given positive integer number n. You should create such strictly increasing sequence of k positive numbers a1, a2, ...…

Description Colossal! — exclaimed Hawk-nose. — A programmer! That's exactly what we are looking for. Arkadi and Boris Strugatsky. Monday starts on Saturday Reading the book "Equations of Mathematical Magic" Roman Oira-Oira and Cristobal Junta fo…

Online Judge:Hdu5155 Label:思维题+容斥,计数Dp 题面: 题目描述 给定一个大小为\(N*M\)的神奇盒子,里面每行每列都至少有一个钻石,问可行的排列方案数.由于答案较大,输出对\(1e9+7\)取模后的结果. 输入 多组数据.每组数据读入两个整数\(N,M\) \(0≤N,M≤50\) 输出 每组数据输出一行表示答案. 样例 Input 1 1 2 2 2 3 Output 1 7 25 Hint There are 7 possible arrangements…

Magic Line 题意 给出n(偶)个整点 整点范围1000,找出一条直线,把n个点分成均等的两部分 分析 因为都是整数,并且范围比较小,所以直接按x排序找到在中间那一部分,并且把中间那一部分的点按照左右点的分布情况,分成两部分即可.如何分呢,因为范围比较小,所以可以找一条斜率特别极限的直线把其分成两部分.这题的关键是要同意直线的形状,要么从左上到右下,要么从左下到右上,混淆就会WA. #include<bits/stdc++.h> using namespace std; #define…

题意:给定三个数,让你放上+-*三种符号,使得他们的值最小. 析:没什么好说的,全算一下就好.肯定用不到加,因为是非负数. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream>…

题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=5176 AX+BY = XY  => (X-B)*(Y-A)= A*B 对A*B因式分解,这里不要乘起来,分A,B因式分解 #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<vector> #inc…

题目链接 题意 :给你第 i 项的值fi,第 j 项的值是 fj 让你求第n项的值,这个数列满足斐波那契的性质,每一项的值是前两项的值得和. 思路 :知道了第 i 项第j项,而且还知道了每个数的范围,二分求第 i+1项,然后根据性质求下去,求到第 j 项的时候看看通过二分求出来的值与给定的第j项的值大小关系,来确定下一次二分的值,输出的时候注意方向. #include <stdio.h> #include <string.h> #include <iostream> #…

题意: 给n和k,让你用不小于 k 个不同的数字构成一个长度为n的序列,使得序列中不同的区间和的数目最小. n,k<=500 k-1个数填一些数字的一正一负,这样有些区间和为0. 剩下的都填0. #include <iostream> #include <cstdlib> #include <cstdio> #include <cstring> #include <string> using namespace std; #define m…