Given a positive integer n, find the least number of perfect square numbers (for example, 1, 4, 9, 16, ...) which sum to n.

For example, given n = 12, return 3 because 12 = 4 + 4 + 4; given n = 13, return 2 because 13 = 4 + 9.

Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

解题技巧：

采用动态规划求解，状态转移方程为：dp[j*j+i]  = min(dp[j*j+i], dp[i]+1)

代码：

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

#define MAX 0x3f3f3f

int numSquares(int n) 
{
	vector<int> dp(n + 1, MAX);
	dp[0] = 0;
	for (int i = 0; i <= n; i++)
	{
		for (int j = 1; j*j + i <= n; j++)
		{
			dp[j*j + i] = min(dp[j*j + i], dp[i] + 1);
		}
	}
	return dp[n];
}

int main()
{
	int n;
	while(cin >> n)
		cout << numSquares(n) << endl;
	return 0;
}