题意:给出一个整数nnn, 找出一个大于等于nnn的最小整数mmm, 使得mmm可以表示为2a3b5c7d2^a3^b5^c7^d2​a​​3​b​​5​c​​7​d​​. 析:预处理出所有形为2a3b5c7d2^a3^b5^c7^d2​a​​3​b​​5​c​​7​d​​即可, 大概只有5000左右个.然后用二分查找就好. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio…

I will show you the most popular board game in the Shanghai Ingress Resistance Team. It all started several months ago. We found out the home address of the enlightened agent Icount2three and decided to draw him out. Millions of missiles were detonat…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5878 题目大意: 给出一个数n ,求一个数X, X>=n. X 满足一个条件 X= 2^a*3^b*5^c*7^d 求靠近n的X值. 解题思路: 打表+二分查找 [切记用 cin cout,都是泪...] AC Code: #include<bits/stdc++.h> using namespace std; ]; int main() { ; long long ans,n,x; ; i…

Problem Description Give you three sequences of numbers A, B, C, then we give you a number X. Now you need to calculate if you can find the three numbers Ai, Bj, Ck, which satisfy the formula Ai+Bj+Ck = X.   Input There are many cases. Every data cas…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2141 题目大意:查找是否又满足条件的x值. 这里简单介绍一个小算法,二分查找. /* x^2+6*x-7==y 输入y 求x 精确度为10^-5 0=<x<=10000 */ #include <iostream> #include <cstdio> using namespace std; int main (void) { double y; while(cin>…

Give you three sequences of numbers A, B, C, then we give you a number X. Now you need to calculate if you can find the three numbers Ai, Bj, Ck, which satisfy the formula Ai+Bj+Ck = X. Input There are many cases. Every data case is described as foll…

题目链接 题意:给定一个数n,求大于n的第一个只包含2357四个因子的数(但是不能不包含其中任意一种),求这个数. 题解:打表+二分即可. #include <iostream> #include <math.h> #include <stdio.h> #include<algorithm> using namespace std; ],tot=; int main() { ; long long a,b,c,d; ;a<=maxn;a*=) ;a*b…

I Count Two Three Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 767    Accepted Submission(s): 403 Problem Description I will show you the most popular board game in the Shanghai Ingress Resis…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4941 解题报告:给你一个n*m的矩阵,矩阵的一些方格中有水果,每个水果有一个能量值,现在有三种操作,第一种是行交换操作,就是把矩阵的两行进行交换,另一种是列交换操作,注意两种操作都要求行或列至少要有一个水果,第三种操作是查找,询问第A行B列的水果的能量值,如果查询的位置没有水果,则输出0. 因为n和m都很大,达到了2*10^9,但水果最多一共只有10^5个,我的做法是直接用结构体存了之后排序,然后m…

<二分查找+双指针+前缀和>解决子数组和排序后的区间和 题目重现: 给你一个数组 nums ,它包含 n 个正整数.你需要计算所有非空连续子数组的和,并将它们按升序排序,得到一个新的包含 n * (n + 1) / 2 个数字的数组. 请你返回在新数组中下标为 left 到 right (下标从 1 开始)的所有数字和(包括左右端点).由于答案可能很大,请你将它对 10^9 + 7 取模后返回. 示例 1:输入:nums = [1,2,3,4], n = 4, left = 1, right…

I Count Two Three 31.1% 1000ms 32768K   I will show you the most popular board game in the Shanghai Ingress Resistance Team. It all started several months ago. We found out the home address of the enlightened agent Icount2three and decided to draw hi…

http://acm.hdu.edu.cn/showproblem.php?pid=4750 题意: 定义f(u,v)为u到v每条路径上的最大边的最小值..现在有一些询问..问f(u,v)>=t的点对有所少对,注意(1,2)和(2,1)是不同的点对 分析: 原来最小生成树有一个很鬼畜的结论,那就是一个图的最小生成树中任意两个点的路径中的最大边一定最小.(妈蛋,完全不知道这个) 然后此题就变得很明朗了,用kruskal算法,加边的时候此边连接的两个集合的路径中的最大边就是这个边,存储下来,询问的时…

题目链接:hdu5878 I Count Two Three 题意:给出一个整数n, 找出一个大于等于n的最小整数m, 使得m可以表示为2^a * 3^b * 5^c * 7^d​​. 题解:打表预处理出所有满足要求的数,排个序然后二分查找解决. #include<cstdio> #include<algorithm> using namespace std; typedef long long ll; const int N = 1e9; ll s[]; ll pow(ll a,…

题目链接:hdu_5878_I Count Two Three 题意: 给你一个n,让你找满足那个式子的不比n小的最小数 题解: 先上个预处理,然后二分查找就行 #include<bits/stdc++.h> #define F(i,a,b) for(int i=a;i<=b;i++) using namespace std; typedef long long ll; ll dt[],ed; ll pow_(ll a,ll k) { ll an=; )an*=a;k>>=,…

Can you find it? Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/10000 K (Java/Others)Total Submission(s): 9180    Accepted Submission(s): 2401 Problem Description Give you three sequences of numbers A, B, C, then we give you a number…

题目链接: Can you find it? Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/10000 K (Java/Others)Total Submission(s): 38634    Accepted Submission(s): 9395 Problem Description Give you three sequences of numbers A, B, C, then we give you a…

OO's Sequence Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 955    Accepted Submission(s): 358 Problem Description OO has got a array A of size n ,defined a function f(l,r) represent the nu…

HDU 2254 奥运(矩阵高速幂+二分等比序列求和) ACM 题目地址:HDU 2254 奥运 题意:  中问题不解释. 分析:  依据floyd的算法,矩阵的k次方表示这个矩阵走了k步.  所以k天后就算矩阵的k次方.  这样就变成:初始矩阵的^[t1,t2]这个区间内的v[v1][v2]的和.  所以就是二分等比序列求和上场的时候了. 跟HDU 1588 Gauss Fibonacci的算法一样. 代码: /* * Author: illuz <iilluzen[at]gmail.com>…

Doubles Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 4266    Accepted Submission(s): 2945 Problem Description As part of an arithmetic competency program, your students will be given randoml…

题目传送门 题意:问区间内x的出现的次数分析:莫队算法:用一个cnt记录x的次数就可以了.还有二分查找的方法 代码: #include <cstdio> #include <algorithm> #include <cmath> #include <cstring> using namespace std; const int MAXN = 1e5 + 10; const int INF = 0x3f3f3f3f; struct Data { int b,…

I Count Two Three 二分查找用lower_bound 这道题用cin,cout会超时... AC代码: /* */ # include <iostream> # include <cstring> # include <string> # include <cstdio> # include <cmath> # include <algorithm> using namespace std; const int TWO…

Vases and Flowers 题目链接 http://acm.hdu.edu.cn/showproblem.php?pid=4614 Problem Description Alice is so popular that she can receive many flowers everyday. She has N vases numbered from 0 to N-1. When she receive some flowers, she will try to put them…

题目链接:hdu_5968_异或密码 题意: 中午,不解释 题解: 前缀处理一下异或值,然后上个二分查找就行了,注意是unsigned long long #include<bits/stdc++.h> #define F(i,a,b) for(int i=a;i<=b;i++) using namespace std; typedef long long ll; ; int t,dp[N],a[N],n,m,tp; struct dt { int val,len; bool opera…

Give you three sequences of numbers A, B, C, then we give you a number X. Now you need to calculate if you can find the three numbers Ai, Bj, Ck, which satisfy the formula Ai+Bj+Ck = X. Input There are many cases. Every data case is described as foll…

Input示例 5 1 8 13 35 77 Output示例 2 8 15 36 80 分析:将所有的只含有2 3 5因子的数打一个表保存在一个数组里,然后二分查找第一个>=数组里的数,输出 #include <bits/stdc++.h> using namespace std; typedef long long LL; #define rep(i,a,n) for(int i = a; i < n; i++) #define repe(i,a,n) for(int i =…

题意: 给定一个素数p(p <= 1e12),问是否存在一对立方差等于p. 分析: 根据平方差公式: 因为p是一个素数, 所以只能拆分成 1*p, 所以 a-b = 1. 然后代入a = b + 1. 求出 3a² + 3a + 1 = p 化简得a(a+1) = (p-1)/3 令(p-1)/3 = T, 问题化为是否存在整数a使得a(a+1) == T, 那么令 t = (int)sqrt(T),只要判定一下t * (t+1) == T ? 即可 另一种做法是打一个a的表(a只要打到1e6)…

解题思路:给出两个数列an,bn,求an和bn中相同元素的个数因为注意到n的取值是0到1000000,所以可以用二分查找来做,因为题目中给出的an,bn,已经是单调递增的,所以不用排序了,对于输入的每一个b[i],查找它在an数列中是否存在.存在返回值为1,不存在返回值为0 CD Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 627…

Problem - 4430 题意是,给出蜡烛的数量,要求求出r和k,r是蜡烛的层数,k是每一层蜡烛数目的底数. 开始的时候,没有看清题目,其实中间的那根蜡烛是可放可不放的.假设放置中间的那根蜡烛,就可以用等比数列求和公式S=(k^(r+1)-1)/(k-1),因为这个公式,对于固定的r,S(k)是单调递增的,然后可以发现,r的范围是相当的小的,最多不会大于40,然后就可以对于每一个r进行一次二分查找,找到k.但是,对于r==1的时候,用这个公式的是会爆龙龙的,而r==1的时候又是直接可以计算出…

题面: 传送门:http://codeforces.com/problemset/problem/475/D Given a sequence of integers a1, -, an and q queries x1, -, xq on it. For each query xi you have to count the number of pairs (l, r) such that 1 ≤ l ≤ r ≤ n and gcd(al, al + 1, -, ar) = xi. 题目大意:…

最新IP地址数据库  来自 qqzeng.com 利用二分逼近法(bisection method) ,每秒300多万, 比较高效! 原来的顺序查找算法 效率比较低 readonly string ipBinaryFilePath = "qqzengipdb.dat"; readonly byte[] dataBuffer, indexBuffer; ]; readonly int dataLength; public IpLocation() { try { FileInfo fil…