Codefroces 280 div2 A. Vanya and Cubes

A. Vanya and Cubes

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Vanya got n cubes. He decided to build a pyramid from them. Vanya wants to build the pyramid as follows: the top level of the pyramid must consist of 1 cube, the second level must consist of 1 + 2 = 3cubes, the third level must have 1 + 2 + 3 = 6 cubes, and so on. Thus, the i-th level of the pyramid must have 1 + 2 + ... + (i - 1) + i cubes.

Vanya wants to know what is the maximum height of the pyramid that he can make using the given cubes.

Input

The first line contains integer n (1 ≤ n ≤ 104) — the number of cubes given to Vanya.

Output

Print the maximum possible height of the pyramid in the single line.

Sample test(s)

input

1

output

1

input

25

output

4

Note

Illustration to the second sample:

很简单的一道打表题，也就是怎么样才能推出来递推公式的关键了，主要就是通过总结了，

第一层1，

第二层1+2，

第三层1+2+3，

，，，

第i层 1+2+3+...+i 

那么sum = 1+1+2+1+2+3+1+2+..+i+..+1+2+..+n;

得到了第i层所需要的方块数和第i+1层所需要的方块数目之间的关系:

a[i+1] = a[i]+i+1;

然后查表，找出上届就可以了。

代码：

# include<cstdio># include<iostream>using namespace std;# define MAX 1234int a[MAX];void dabiao(){    a[1] = 1;    for ( int i = 1;i < 1000;i++ )        {            a[i+1] = a[i]+i+1;        }}int main(void){    dabiao();    int n;    while ( cin>>n )        {            int ans = 0;            int i = 1;            while ( n-a[i]>=0 )                {                    n = n-a[i];                    ans = i;                    i++;                }                cout<<ans<<endl;        }    return 0;}