2013
11-10

# Lagrange’s Four-Square Theorem

The fact that any positive integer has a representation as the sum of at most four positive squares (i.e. squares of positive integers) is known as Lagrange’s Four-Square Theorem. The first published proof of the theorem was given by Joseph-Louis Lagrange in 1770. Your mission however is not to explain the original proof nor to discover a new proof but to show that the theorem holds for some specific numbers by counting how many such possible representations there are.

For a given positive integer n, you should report the number of all representations of n as the sum of at most four positive squares. The order of addition does not matter, e.g. you should consider 4^2 + 3^2 and 3^2 + 4^2 are the same representation.

For example, let’s check the case of 25. This integer has just three representations 1^2+2^2+2^2+4^2, 3^2 + 4^2, and 5^2. Thus you should report 3 in this case. Be careful not to count 4^2 + 3^2 and 3^2 + 4^2 separately.

The input is composed of at most 255 lines, each containing a single positive integer less than 2^15, followed by a line containing a single zero. The last line is not a part of the input data.

The output should be composed of lines, each containing a single integer. No other characters should appear in the output.

The output integer corresponding to the input integer n is the number of all representations of n as the sum of at most four positive squares.

1
25
2003
211
20007
0

1
3
48
7
738


/* @author: */
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);

int n, s, t1, t2, t;
while( sc.hasNext() )
{
n=sc.nextInt();
if( n == 0 ) break;
s = 0;
for(int i=(int)Math.sqrt( n/4 ); i*i<=n; i++ )
{
t1=n-i*i;
for(int j=(int)Math.sqrt( t1/3 ); j<=i && j*j<=t1; j++ )
{
t2=t1-j*j;
for(int k=(int)Math.sqrt(t2/2) ; k<=j && k*k<=t2; k++ )
{
t = (int)Math.sqrt( t2 - k*k );
if( t <= k && t*t == t2 - k*k )
s++;
}
}
}
System.out.printf( "%d\n", s );
}
}
}

1. 思路二可以用一个长度为k的队列来实现，入队后判断下队尾元素的next指针是否为空，若为空，则出队指针即为所求。

2. 换句话说，A[k/2-1]不可能大于两数组合并之后的第k小值，所以我们可以将其抛弃。
应该是，不可能小于合并后的第K小值吧