class Solution {
public:int numSquares(int n) {vector<int> f(n + 1);for (int i = 1; i <= n; i++) {int minn = INT_MAX;for (int j = 1; j * j <= i; j++) {minn = min(minn, f[i - j * j]);   //上一次的 & 当前数可以找到一个新的更大的平方}f[i] = minn + 1;     }return f[n];}
};