2020声网日常实习C++笔试题:求数组的最大公约数

题目描述:
Let’s consider all in the range from 1 to n(inclusive).
Among all pairs of distinct integers in this range, find the maximum possible greatest common divisor of integers in pair. Formally, find the maximum value of gcd(a ,b), where 1 ≤ \leq a ≤ \leq b ≤ \leq n.
The greatest common divisor, gcd(a, b), of two positive integers a and b is the biggest integer that is a divisor of both a and b.

输入描述:
The first line contains a single integer t (1 ≤ \leq t ≤ \leq 100) - the number of test cases. The description of the test cases follows.The only one line of each test case contains a single integer (2 ≤ \leq n ≤ \leq 106) .

输出描述:
For each test case, output the maximum value of gcd(a, b) among all 1 ≤ \leq a ≤ \leq b ≤ \leq n.

首先给出最终的AC通过结果代码:

#include <iostream>
using namespace std;
int main(){
    
    
	int NumOfInput; //输入的个数
	cin >> NumOfInput;
	
	while(NumOfInput--){
    
    
		int MaxNum;
		cin >> MaxNum; //n的最大值
		
		if(MaxNum % 2 == 0)
			cout << MaxNum / 2 << endl;	
		else
			cout << (MaxNum-1) /2 << endl;
	}	
	return 0;	
}

分析:

  • 题目看起来比较复杂,总结为一句话“求最大公约数的最大值”。 最常用的求两个数的最大公约数的求法为辗转相除法,其实现代码如下:
int GreaComDiv(int a, int b){
    
    
	int c = b;	
	while(a%b != 0){
    
    
		c = a % b;
		a = b;
		b = c;
	} 
	return c;
}
  • 求是最大公约数的最大值,即转为求2与n或者n-1的最大公约数,理解了这个之后接下来就简单啦,直接往上套就可以了!

总结:

  • 求最大公约数的最大值,可以把问题转化为为求最小值和最大值的最大公约数。

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转载自blog.csdn.net/zxxkkkk/article/details/109242739