简单二分法

点击打开链接
#include<iostream>

#include<algorithm>
using namespace std;
#define ll long long
int num[50005];
int main()
{
	ll n,k;
	cin>>n>>k;
	for (int i=1;i<=k;i++)
	{
		cin>>num[i];
	}
	sort(num+1,num+k+1);
	bool flag=true;
	ll i=1,j=k;
	while(i<j)
	{
	    if (num[i]+num[j]==n)
		{
			flag=false;
			cout<<num[i++]<<" "<<num[j--]<<endl;
		}
		else if (num[i]+num[j]<n)
		i++;
		else 
		j--;
	}
	if (flag==true)
	cout<<"No Solution"<<endl;
	return 0;
}

点击打开链接

#include <iostream>

#include <cmath>

#include<algorithm>
using namespace std;
int main()
{
    long long n;
    cin>>n;
    int i=0,j=sqrt(n),len=sqrt(n/2);
    bool flag=true;
    while (i<j)
    {	
        if ((i*i+j*j)==n)
    	{
    		flag=false;
    		cout<<i<<" "<<j<<endl;
    		i++;j--;
    	}
    	else if ((i*i+j*j)<n)
    	i++;
    	else
    	j--;
    }
    if (2*(len*len)==n)
    {
    	flag=false;
    	cout<<len<<" "<<len<<endl;
    }
    if (flag)
    cout<<"No Solution"<<endl;
    return 0;
}

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转载自blog.csdn.net/qq_40911499/article/details/79843855