ural 1073. Square Country

1073. Square Country

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	output
344	3

Problem Author: Stanislav Vasilyev
Problem Source: Ural State Univerisity Personal Contest Online February'2001 Students Session

Tags: dynamic programming  (hide tags for unsolved problems)

Difficulty: 161

 

题意：给出x，问要多少个完全平方数相加才能得到x。

分析：dp可以解决这个问题。

dp[i]表示i最少要多少个来组成、

dp[i] = min(d[i - j*j] + 1)

 

但是这题有更简单的做法。

根据定理，仍和正整数可以有四个完全平方数组成。

所以说最多四个数。

这样直接暴力即可。

当然，我是弱智，做的时候没有想到。

 

 /**
 Create By yzx - stupidboy
 */
 #include <cstdio>
 #include <cstring>
 #include <cstdlib>
 #include <cmath>
 #include <deque>
 #include <vector>
 #include <queue>
 #include <iostream>
 #include <algorithm>
 #include <map>
 #include <set>
 #include <ctime>
 #include <iomanip>
 using namespace std;
 typedef long long LL;
 typedef double DB;
 #define MIT (2147483647)
 #define INF (1000000001)
 #define MLL (1000000000000000001LL)
 #define sz(x) ((int) (x).size())
 #define clr(x, y) memset(x, y, sizeof(x))
 #define puf push_front
 #define pub push_back
 #define pof pop_front
 #define pob pop_back
 #define ft first
 #define sd second
 #define mk make_pair

 inline int Getint()
 {
     int Ret = ;
     char Ch = ' ';
     bool Flag = ;
     while(!(Ch >= '' && Ch <= ''))
     {
         if(Ch == '-') Flag ^= ;
         Ch = getchar();
     }
     while(Ch >= '' && Ch <= '')
     {
         Ret = Ret *  + Ch - '';
         Ch = getchar();
     }
     return Flag ? -Ret : Ret;
 }

 const int N = ;
 int n;
 int dp[N];

 inline void Input()
 {
     cin >> n;
 }

 inline void Solve()
 {
     dp[] = , dp[] = ;
     for(int i = ; i <= n; i++)
     {
         dp[i] = INF;
         for(int j = ; j <= i / j; j++)
             if(dp[i] > dp[i - j * j] + )
                 dp[i] = dp[i - j * j] + ;
     }
     cout << dp[n] << endl;
 }

 int main()
 {
     freopen("e.in", "r", stdin);
     Input();
     Solve();
     return ;
 }