搜索 HOJ 1281 Lagrange\'s Four-Square Theorem

Lagrange's Four-Square Theorem

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	Source : ACM ICPC Aizu(Japan) Regional Contest 2003
	Time limit : 1 sec		Memory limit : 32 M

Submitted : 97, Accepted : 57The 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.
Input[/b]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.
Output[/b]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.
Sample Input[/b]1
25
2003
211
20007
0
Sample Output[/b]1
3
48
7
738


题意：任意一个数都能被不多于4个的数字的平方相加表示，问一个数有多少种表示方法。
思路：枚举吧，枚举第一个数，第二个数，第三个数，第四个数，很简单是吧。。。。
代码：#include<cstdio>#include<iostream>#include<algorithm>#include<cstring>#include<string.h>#include<math.h>#include<string.h>using namespace std;const int maxn = (1<<15)+10;
int ans[maxn];

int main(){ for (int i = 1 ; i*i < maxn ; ++i) { int a = i*i; ++ans[i*i]; for (int j = i ; a + j*j < maxn ; ++j) { int b = a+j*j; ++ans[b]; for (int k = j ; b + k*k < maxn ; ++k) { int c = b+k*k; ++ans[c]; for (int l = k ; c+l*l < maxn ; ++l) ++ans[c+l*l]; } } } int n; while (scanf("%d",&n),n) { printf("%d\n",ans
); }}