HDU6025 Coprime Sequence —— 前缀和 & 后缀和
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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 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 can try to maximize the GCD of these integers by removing exactly one integer. Now given a sequence, please maximize the GCD of its elements.
``Coprime Sequence'' is easy to find because of its restriction. But we can try to maximize the GCD of these integers by removing exactly one integer. Now given a sequence, please maximize the GCD of its elements.
Input
The first line of the input contains an integer T(1≤T≤10) , denoting the number of test cases.
In each test case, there is an integern(3≤n≤100000) in the first line, denoting the number of integers in the sequence.
Then the following line consists ofn integers a1,a2,...,an(1≤ai≤109) , denoting the elements in the sequence.
In each test case, there is an integer
Then the following line consists of
Output
For each test case, print a single line containing a single integer, denoting the maximum GCD.
Sample Input
331 1 152 2 2 3 241 2 4 8
Sample Output
122
题解:
l[i]为前i个数的gcd, r[i]为后i个数的gcd。
假设被删除的数的下标为i, 则 删除该数后的gcd为: gcd(l[i-1], r[i+1]), 枚举i,取最大值。
学习之处:
当提到在序列里删除一段连续的数时,可以用前缀和+后缀和。
例如:http://blog.csdn.net/dolfamingo/article/details/71001021
代码如下:
#include<bits/stdc++.h>using namespace std;typedef long long LL;const double eps = 1e-6;const int INF = 2e9;const LL LNF = 9e18;const int mod = 1e9+7;const int maxn = 1e5+10;int n;int a[maxn], l[maxn], r[maxn];int gcd(int a, int b){ return b==0?a:(gcd(b,a%b));}void solve(){ scanf("%d",&n); for(int i = 1; i<=n; i++) scanf("%d",&a[i]); l[1] = a[1]; r[n] = a[n]; for(int i = 2; i<=n; i++) l[i] = gcd(l[i-1], a[i]); for(int i = n-1; i>=1; i--) r[i] = gcd(r[i+1], a[i]); int ans = 1; l[0] = a[2]; r[n+1] = a[n-1]; for(int i = 1; i<=n; i++) ans = max(ans, gcd(l[i-1], r[i+1]) ); cout<<ans<<endl;}int main(){ int T; scanf("%d",&T); while(T--) { solve(); } return 0;}
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