题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=6025 Coprime Sequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 666    Accepted Submission(s): 336 Problem Description Do you know what is…

Do you know what is called ``Coprime Sequence''? That is a sequence consists of nnpositive integers, and the GCD (Greatest Common Divisor) of them is equal to 1. ``Coprime Sequence'' is easy to find because of its restriction. But we can try to maxim…

题意:去除数列中的一个数字,使去除后数列中所有数字的gcd尽可能大. 分析:这个题所谓的Coprime Sequence,就是个例子而已嘛,题目中没有任何语句说明给定的数列所有数字gcd一定为1→_→,队友这样解读题目导致我们队的思路完全想歪了,当时真的脑子发懵,为啥没再读一遍题= = 1.求出一个数列的gcd跟计算顺序没关系. 2.如果求去掉下标为i的数字后整个数列的gcd,直接将该数字前的所有数字的gcd(prefix[i-1])和该数字后所有数字的gcd(suffix[i+1])再求一下g…

Description Do you know what is called ``Coprime Sequence''? That is a sequence consists of $n$ positive integers, and the GCD (Greatest Common Divisor) of them is equal to 1. ``Coprime Sequence'' is easy to find because of its restriction. But we ca…

枚举,预处理. 预处理前缀$gcd$与后缀$gcd$,枚举删哪一个即可. #include <bits/stdc++.h> using namespace std; int T,n; ]; ]; ]; long long gcd(long long a,long long b) { ) return a; return gcd(b,a%b); } int main() { scanf("%d",&T); while(T--) { scanf("%d&quo…

Problem Description Generally speaking, there are a lot of problems about strings processing. Now you encounter another such problem. If you get two strings, such as “asdf” and “sdfg”, the result of the addition between them is “asdfg”, for “sdf” is…

题意:给出一串数列,这串数列的gcd为1,要求取出一个数使取出后的数列gcd最大. 题解:可以通过对数列进行预处理,求出从下标为1开始的数对于前面的数的gcd(数组从下标0开始),称为前缀gcd,再以类似的方式求出后缀gcd,然后从第一个数开始枚举取出后的gcd(这个数的前缀gcd与后缀gcd的gcd).找出最大的gcd数即可. 代码如下: #include<cstdio> #include<cstring> #include<algorithm> using name…

大致题意: 去掉一个元素能使这个数列的GCD最大为多少 分析: 我们求一个数列的GCD,是先求前两个元素的GCD,然后将这个GCD值在与下一个元素进行GCD运算.由此可知进行GCD运算的顺序对最终的结果是没有影响的.我们再看看数列的长度范围,小于100000.那我们就枚举去掉的那个元素,那么去掉元素后的数列的GCD即这个元素前面一段数列的GCD再与这个元素后面一段数列的GCD进行GCD运算.所以我们需要两个数组分别记录前缀GCD和后缀GCD,这两个数组都可以通过O(n)算法算出来. #inclu…

HDU6025 Coprime Sequence 处理出数列的 \(gcd\) 前缀和后缀,删除一个数后的 \(gcd\) 为其前缀和后缀的 \(gcd\) . 遍历数列取 \(max\) 即为答案. 时间复杂度为 \(O(n)\) . #include<bits/stdc++.h> using namespace std; const int maxn = 100005; int a[maxn], head[maxn], tail[maxn]; int t, n; int gcd(int a…

C HDU - 6025 [题意]:去除数列中的一个数字,使去除后的数列中所有数字的gcd尽可能大. [分析]: 数组prefixgcd[],对于prefixgcd[i]=g,g为a[0]-a[i]的GCD,称为前缀GCD. 数组suffixgcd[],对于suffixgcd[i]=g,g为a[i]-a[n-1]的GCD,称为后缀GCD. 有了这两个GCD值的数组,那么去掉a[i]的GCD为gcd(prefixgcd[i - 1], suffixgcd[i + 1]),从中找出最大值即可. [代…