（动态规划）ural 1073 Square Country

Square Country

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status Practice URAL 1073

Description

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 positive integer 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.

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.

Input

The only line contains a positive integer N ≤ 60 000 , that is a number of quadrics that the citizen has invested.

Output

The only line contains a number of certificates that he has gotten.

Sample Input

inputoutput

344

3

 题意：

 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.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.

输入包含一个自然数N,N<=60000,表示他能买多少方土地

思路分析：用 DP 的知识解决问题

数组 a 存储的是最少有多少个完全平方数

已知  a[0]=0; a[1]=1;可推得：

a[2]=a[1]+1 得2;

a[3]=a[2]+1 得3;

a[4]=a[0+2^2] 得1;

a[5]=a[4]+1 得2;

a[6]=a[5]+1 得3;

dp方程为：a[i]=min(a[i],a[i-j*j]+1);

源程序为：

#include<iostream>#include<string.h>#define INF 0x3f3f3f3f#include<algorithm>using namespace std;int a[60010];int main(){    int n;    memset(a,INF,sizeof(a));    a[0]=0;    a[1]=1;    for(int i=2;i<=60000;i++){        for(int j=1;j<=245;j++){            if(i>=j*j)         a[i]=min(a[i],a[i-j*j]+1);            else    break;        }    }    cin>>n;    cout<<a[n]<<endl;    return 0;}

注意：min和max是标准库<algorithm>里面的，想直接用这两个函数时需写头文件#include<algorithm>