HDU6025 Coprime Sequence —— 前缀和 & 后缀和
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
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.
denoting the number of test cases.
In each test case, there is an integer n(3≤n≤100000) in
the first line, denoting the number of integers in the sequence.
Then the following line consists of n integers a1,a2,...,an(1≤ai≤109),
denoting the elements in the sequence.
3
3
1 1 1
5
2 2 2 3 2
4
1 2 4 8
1
2
2
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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