1221. 四平方和(二分查找)

二分搜索的思想来优化 

#include 
#include 
using namespace std;
const int N = 10005;
int k[N];
int binary_search(int x)
{
	int l = 0, r = 10000;
	while (l < r)
	{
		int mid = l + r >> 1;
		if (mid * mid < x)
			l = mid + 1;
		else if (mid * mid == x)
			return mid;
		else if (mid * mid > x)
			r = mid;
	}
	return -1;
}
int main()
{
	int n;
	cin >> n;
	//n^3 logn
	int kk = 0;
	for(int i=0;i*i<=n;i++)
		for(int j=0;j*j<=n;j++)
			for (int k = 0;k * k <= n;k++)
			{
				int x = n - i * i - j * j - k * k;
				int kk = binary_search(x);
				if (kk == -1) continue;
				else
				{
					cout << i << ' ' << j << ' ' << k << ' ' << kk;
					exit(0);
				}
			}
	return 0;
}

#include 
#include 
#include 
using namespace std;
const int N = 2500100;    //记得开大一点，否则答案会出错
 //因为按照data排序后，可能0 4000 这个数在很后面，数组越界+wa
struct p
{
	int x, y;
	int data;
	bool operator <(const p& b)const
	{
		if (data == b.data)
			return x < b.x ? true : false;
		return data < b.data ? true : false;
	}
}k[N];
int cnt = 0;
//返回下标
int Binary_search(int c)
{
	int l = 0, r = cnt - 1;
	while (l < r)
	{
		int mid = l + r >> 1;
		if (k[mid].data < c)
			l = mid + 1;
		if (k[mid].data >= c)
			r = mid;
	}
	return k[l].data == c ? l : -1;
}
int main()
{
	int n;
	cin >> n;
	for(int i=0;i * i <=n;i++)
		for (int j = i;j * j+i*i<= n;j++)
		{
			k[cnt++] = { i,j,i * i + j * j };
		}
	sort(k, k + cnt);
	for (int i = 0;i * i <= n;i++)
		for (int j = 0;j * j+ i*i <= n;j++)
		{
			int c = n - i * i - j * j;
			int l = 0, r = cnt - 1;
			while (l < r)
			{
				int mid = l + r >> 1;
				if (k[mid].data < c)
					l = mid + 1;
				if (k[mid].data >= c)
					r = mid;
			}
			if (k[l].data != c) continue;
			int x = k[l].x;
			int y = k[l].y;
			cout << i << ' ' << j << ' ' << x << ' ' << y << ' ';
			exit(0);
		}
}