题意：
A square problem Time Limit: 3 second Memory Limit: 1000K
There live square people in a square country. Everything in this country is square also. Thus, the Square Parliament has passed a law about a land. According to the law each citizen of the country has a right to buy land. A land is sold in squares, surely. Moreover, a length of a square side must be a natural amount of meters. Buying a square of land with a side a one pays a2 quadrics (a local currency) and gets a square certificate of a landowner.
在一个正方形的国度里住着正方形的人.在这个国家里,所有的东西都是正方形的.该国的国会通过了一项关于土地的法律,依照法律,该国的国民有买土地的权利,当然,土地的买卖也是按照正方形进行.而且,买卖的土地的边长必须是整米数,每买一块土地,必须付款(用当地的钱币),每买一块地,买主会得到一份土地所有者的证明.
One citizen of the country has decided to invest all of his N quadrics into the land. He can, surely, do it, buying square pieces 1 × 1 meters. At the same time the citizen has requested to minimize an amount of pieces he buys: "It will be easier for me to pay taxes," - he has said. He has bought the land successfully. Your task is to find out a number of certificates he has gotten.
一个市民打算把他的钱投资到土地上,因为都只能买边长为整数的正方形地,他希望土地的块数最小.他认为:"这使我在交税时,更方便",他终于购地成功. 你的任务是找出他购地的块数,以便发给他地主证书.
Input The only line contains a natural number N<=60 000 - that is a number of quadrics that the citizen has invested.
输入包含一个自然数N,N<=60000,表示他能买多少方土地
Output The only line contains a number of certificates that he has gotten.
输出他得到的土地块数.
思路：
完全背包问题。容量是N,单位物品为i*i(i从1到sqrt(N))，转移方程式为f[j]=min(f[j],f[j-i*i]+1)，注意初始化。
（关于完全背包问题，建议看一下《背包九讲》）
代码：
Square Country
#include <iostream>
using namespace std;
const int N=60000;
int f[N];
int main()
{
int n;
cin>>n;
for(int i=0;i<=n;i++) f[i]=i;//初始化
for(int i=1;i*i<=n;i++)
{
for(int j=i*i;j<=n;j++)
{
f[j]=min(f[j],f[j-i*i]+1);
}
}
cout<<f[n]<<endl;
return 0;
}