ACM第二次比賽（ C )

標籤：

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Description

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 ≤ 10^{4) — the number of cubes given to Vanya.}

Output

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

Sample Input

Input

1

Output

1

Input

25

Output

4

 

 

程式分析：此題的考點還是在於累加。第一次是1，第二次就是1+2，第三次就是1+（1+2）+3如此迴圈，所以這也要求了我們應該把每一次答案放在一個數裡，對於這個題目我們用數組更能夠解決問題，在用一個for迴圈把要加的數加進來。但是應該注意的是最後我們sum-n>0的時候，我們是加到超過了n的，所以j應該減1.但是如果sum==0,則j的值就是我們要得到的值。

程式碼：

#include<cstdio>#include<iostream>using namespace std;int main( ){int a[200]={0};int i,n,j,sum=0;a[1]=1;for(i=2;i<200;i++){for(j=0;j<=i;j++)a[i]+=j;}scanf("%d",&n);for(j=1;j<=i;j++){sum+=a[j];if(sum>n){printf("%d\n",j-1);break;}else if(sum==n){printf("%d\n",j);break;}}return 0;}

ACM第二次比賽（ C )