Acwing1221. Sum of Four Squares (Three Cycle Optimization)

The four-square sum theorem, also known as Lagrange's theorem:

Every positive integer can be represented as the sum of squares of at most 4 positive integers.

If 0 is included, it can be expressed as the sum of squares of 4 numbers.

for example:

5=02+02+12+22
7=12+12+12+22

For a given positive integer, there may be multiple representations of the sum of squares.

You are asked to sort 4 numbers:

0≤a≤b≤c≤d

And arrange all possible representations in ascending order according to a, b, c, d as the joint primary key, and finally output the first representation.

input format

Enter a positive integer N.

output format

Output 4 non-negative integers, sorted from small to large, separated by spaces.

data range

0<N<5∗106

Input sample:

5

Sample output:

0 0 1 2

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>

using namespace std;

int n,i,j,k,l,a,b,c;

int main(){
    cin >> n;
    for(a = i * i; a * 4 <= n; ++ i, a = i * i)
        for(j = i, b = j * j; b * 3 <= n - a; ++ j, b = j * j)
            for(k = j, c = k * k; c * 2 <= n - a - b; ++ k, c = k * k){
                l=sqrt(n - a - b - c);
                if(a + b + c + l * l == n){
                    cout << i << ' ' << j << ' ' << k << ' ' << l;
                    return 0;
                }
            }
    return 0;
}

 When the first number is violent, because i<=j<=k<=l, so i<=n/4;j<=(ni*i)/3;k<=(ni*ij*j)/2

Think about it again, the program will take a lot of time to calculate the multiplication, so you can define, a=i*i;b=j*j;c=k*k.

Final optimization (the code changes greatly, so it is highlighted with no spaces)

Just AC! ! ! (Although it may happen to be stuck, just pay it a few more times)