1221. 四平方和(非暴力法)

**思路：**先枚举两个数然后将两个数的平方和保存起来放入结构体数组中，再枚举另外两个数，利用二分法判断这两个数的平方和是否在刚刚的数组中

# include<iostream>
# include<cstdio>
# include<cstring>
# include<algorithm>
# include<cmath>
using namespace std;

const int N = 2500010;

struct sum
{
    
    
    int s,c,d;
    bool operator<(const sum& t)const//重载运算符<使得sort可以将结构体的数按照大小排列
    {
    
    
        if(s != t.s)return s < t.s;
        if(c != t.c)return c < t.c;
        return d < t.d;
    }
}sum[N];
int n,m;

int main()
{
    
    
    cin >> n;
    for(int c = 0;c * c <= n;c++)
    {
    
    
        for(int d = c;c * c + d * d <= n;d++)
        {
    
    
            sum[m++] = {
    
    c * c + d * d,c,d};      
        }
    }
    sort(sum,sum + m);
    for(int a = 0;a * a <= n;a ++)
    {
    
    
        for(int b = a ;a * a + b * b <= n;b++)
        {
    
    
            int t = n - a * a - b * b;
            int l = 0,r = m - 1;
            while(l < r)//利用二分判断是否存在一个和t相等的数
            {
    
    
                int mid = (l + r) / 2;
                if(sum[mid].s >= t)
                {
    
    
                    r = mid;
                }
                else
                {
    
    
                    l = mid + 1;
                }
            }
            if(sum[l].s == t)
            {
    
    
                printf("%d %d %d %d\n",a,b,sum[l].c,sum[l].d);
                return 0;
            }
        }
    }
}