HDU6025 Coprime Sequence【前缀GCD+后缀GCD】
2017-05-25 00:17
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Coprime Sequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 1415 Accepted Submission(s): 690
[align=left]Problem Description[/align]
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.
[align=left]Input[/align]
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 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.
[align=left]Output[/align]
For each test case, print a single line containing a single integer, denoting the maximum GCD.
[align=left]Sample Input[/align]
3
3
1 1 1
5
2 2 2 3 2
4
1 2 4 8
[align=left]Sample Output[/align]
1
2
2
[align=left]Source[/align]
2017中国大学生程序设计竞赛 - 女生专场
问题链接:HDU6025 Coprime Sequence。
问题简述:
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。
问题分析:
这是一个计算GCD的问题。
程序说明:
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。
题记:(略)
AC的C++语言程序如下:
/* HDU6025 Coprime Sequence */ #include <iostream> using namespace std; const int N = 100000; int a , prefixgcd , suffixgcd ; inline int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); } int main() { int t, n; cin >> t; while(t--) { cin >> n; for(int i=0; i<n; i++) cin >> a[i]; // 计算前缀GCD prefixgcd[0] = a[0]; for(int i=1; i<n; i++) prefixgcd[i] = gcd(prefixgcd[i - 1], a[i]); // 计算后缀GCD suffixgcd[n - 1] = a[n - 1]; for(int i=n-2; i>=0; i--) suffixgcd[i] = gcd(a[i], suffixgcd[i + 1]); int ans = max(suffixgcd[1], prefixgcd[n - 2]); for(int i=1; i<n - 1; i++) ans = max(ans, gcd(prefixgcd[i - 1], suffixgcd[i + 1])); cout << ans << endl; } return 0; }
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