Perfect Squares

问题：

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

样例

Given n = 12, return 3 because 12 = 4 + 4 + 4
Given n = 13, return 2 because 13 = 4 + 9

这个题目我们可以使用dp（动态规划来实现）：

#include<iostream>#include <cmath>#include <algorithm>using namespace std;class Solution {public:/*** @param n a positive integer* @return an integer*/int numSquares(int n) {// Write your code hereint *dp = new int[n + 1];for (int i = 0; i < n + 1;i++){dp[i] = pow(2, 31) - 1;//将所有非平方数的结果置最大，保证之后比较的时候不被选中}//将所有平方数 置为1for (int i = 0; i*i < n + 1; i++){dp[i*i] = 1;}// 从小到大找任意数afor (int a = 0; a <= n; a++){// 从小到大找平方数bｘbfor (int b = 0; a + b * b <= n; b++){// 因为a+b*b可能本身就是平方数，所以我们要取两个中较小的dp[a + b * b] = min(dp[a] + 1, dp[a + b * b]);}}return dp[n];}};int main(){int n = 0;cin >> n;Solution s;cout << s.numSquares(n) << endl;return 0;}