CodeforcesRound#280(Div.2)A_html/css

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CodeforcesRound#280(Div.2)A_html/css

CodeforcesRound#280(Div.2)A_html/css_WEB-ITnose:題目： 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 d

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導讀CodeforcesRound#280(Div.2)A_html/css_WEB-ITnose:題目： 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 d

題目：

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

output

Note

Illustration to the second sample:

沒啥好說的。模擬一下就可以了，還wa一發(fā)不應該..

#include #include #include #include #include #include #include #include #include #include #include #include #include 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 3.141592653579323846233832795#define HALF_PI 1.57079632679466192313216916398#define MAXN 100000#define MAXM 10000#define MOD 1000000007typedef long long LL;const double maxdouble = numeric_limits::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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CodeforcesRound#280(Div.2)A_html/css

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