Codeforces Round #280 (Div. 2) A_html/css_WEB-ITnose

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发布: 2016-06-24 11:53:09
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题目:

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?=?3 cubes, 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

output

input

25
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output

Note

Illustration to the second sample:

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div+css3阶梯分页样式
div+css3阶梯分页样式

div+css3阶梯分页样式

div+css3阶梯分页样式 84
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没啥好说的。模拟一下就可以了,还wa一发不应该..


#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <vector>#include <queue>#include <stack>#include <cassert>#include <algorithm>#include <cmath>#include <set>#include <limits>#include <map>using namespace std;#define MIN(a, b) ((a) < (b) ? (a) : (b))#define MAX(a, b) ((a) > (b) ? (a) : (b))#define F(i, n) for(int (i)=0;(i)<(n);++(i))#define REP(i, s, t) for(int (i)=(s);(i)<=(t);++(i))#define UREP(i, s, t) for(int (i)=(s);(i)>=(t);--(i))#define REPOK(i, s, t, o) for(int (i)=(s);(i)<=(t) && (o);++(i))#define MEM0(addr) memset((addr), 0, sizeof((addr)))#define PI 3.1415926535897932384626433832795#define HALF_PI 1.5707963267948966192313216916398#define MAXN 100000#define MAXM 10000#define MOD 1000000007typedef long long LL;const double maxdouble = numeric_limits<double>::max();const double eps = 1e-10;const int INF = 0x7FFFFFFF;int main(){    int n;    cin >> n;    int ans = 0;    int base = 1;    while (n >= base)    {        ++ans;        n -= base;        base += ans+1;    }    cout << ans;    return 0;}
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