URAL1073——DP——Square Country

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

input	output
344	3

 大意：你现在有n块土地，要把它分成几块正方形，问最少的正方形数目，原本以为可以贪心做，倒着过来，不过12(4+4+4 = 3) 9+1+1+1 = 4这组数据有问题

动态转移方程  dp[i] = min(dp[i],dp[i-j*j)+1)     

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
int main()
{
    int dp[60010];
    int n;
    while(~scanf("%d",&n)){
        memset(dp,0,sizeof(dp));
        for(int i = 1; i <= n ;i++){
            dp[i] = dp[i-1] + 1;
            for(int j = 1; j <= sqrt((double)n+0.5); j++)
                if(j*j <= i)
                dp[i] = min(dp[i],dp[i-j*j] + 1);
        }
    printf("%d
",dp[n]);
    }
    return 0;
}