POJ1078题解(dfs搜索乘积项)

题意描述:

给两个整数是否可以分解为1-n的整数乘积,且乘积项各不相同,如果是这样的话输出二者最大值,或较小的数不能分解也输出最大值,否则输出较大的那个。

代码:

#include<iostream>
using namespace std;
#include<vector>
#include<algorithm>
int flag[101] = {
    
     0 };
vector<int>temp;
void dfs(int n,vector<vector<int>> &a)
{
    
    	
	if (n == 1)
	{
    
    	
		/*for (int k = 0; k < temp.size()-1; k++)
		{
			if (temp[k] > temp[k + 1])
				return;
		}*/
		a.push_back(temp);
		return;
	}
	for (int i = 2; i <= 100; i++)
	{
    
    
		if (n % i == 0&&flag[i]==0)
		{
    
    	
			flag[i] = 1;
			temp.push_back(i);
			dfs(n / i,a);
			temp.pop_back();
			flag[i] = 0;
		}
	}
}
bool judge(vector<int>a, vector<int>b)
{
    
    
	for (int i = 0; i < a.size(); i++)
	{
    
    
		for (int j = 0; j < b.size(); j++)
		{
    
    
			if (a[i] == b[j])
				return false;
		}
	}
	return true;
}
int main()
{
    
    	
	int a, b;
	while (cin >> a >> b)
	{
    
    	
		vector<vector<int>>num1;
		vector<vector<int>>num2;
		int a1 = 0, b1 = 0;
		memset(flag, 0, sizeof(flag));
		a1 = max(a, b);
		b1 = min(a, b);
		dfs(a1, num1);
		dfs(b1, num2); int flag = 0;
		for (int i = 0; i < num1.size(); i++)
		{
    
    
			for (int j = 0; j < num2.size(); j++)
			{
    
    
				if (judge(num1[i], num2[j]))
				{
    
    
					flag = 1;
					break;
				}
			}
			if (flag)
				break;
		}
		if (flag||num2.size()==0)
			cout << a1<<endl;
		else
			cout << b1<<endl;
	}
	return 0;
}

猜你喜欢

转载自blog.csdn.net/stn54999/article/details/114036303