【Ural1073】 Square Country——完全背包

Time limit: 1.0 second	Memory limit: 64 MB

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

344

output

3

题意：每一条路的价值是路长的平方，最小的路长为1，现在有n元钱，问最少能买多少条路，（n元全部花费）
思路：对于n元钱，可以买的最长的路为 ⌊n−√⌋ ，所以枚举路的价值，则 Dp[j]=min(Dp[j−i∗i]+1，Dp[j]) ，基础的完全背包

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>

using namespace std;

const int Max = 61000;

const int INF = 0x3f3f3f3f;

int Dp[Max];

int main()
{

    int n;

    while(~scanf("%d",&n))
    {
        memset(Dp,INF,sizeof(Dp));//求的为最小值，所以初始化为INF

        Dp[0] = 0;

        for(int i = 1;i*i<=n;i++)
        {
            for(int j = i*i;j<=n;j++)
            {
                Dp[j]  =min(Dp[j],Dp[j-i*i]+1);
            }
        }

        cout<<Dp[n]<<endl;
    }
    return 0;
}