HDU6025 Coprime Sequence【前缀GCD+后缀GCD】

Coprime Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1415    Accepted Submission(s): 690


Problem DescriptionDo 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. 
InputThe 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. 
OutputFor each test case, print a single line containing a single integer, denoting the maximum GCD. 
Sample Input3 3 1 1 1 5 2 2 2 3 2 4 1 2 4 8 
Sample Output1 2 2 
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。

C++语言程序如下：

/* HDU6025 Coprime Sequence */ #include using namespace std; const int N = 100000; int a[N], prefixgcd[N], suffixgcd[N]; inline int gcd(int a, int b) { //inline（修饰词，告诉编译器怎样去编译这段程序，c++特有） return b == 0 ? a : gcd(b, a % b); } //辗转相除法（还有递归） int main() { int t, n; cin >> t; while(t--) { cin >> n; //输入流输入数据 for(int i=0; i> a[i]; //把数放入数组之中 // 计算前缀GCD prefixgcd[0] = a[0]; for(int i=1; i=0; i--) suffixgcd[i] = gcd(a[i], suffixgcd[i + 1]); //从前面算一次gcd，再从后面算一次gcd int ans = max(suffixgcd[1], prefixgcd[n - 2]); //去掉两头的 for(int i=1; i