Codeforces Round #280 (Div. 2)-A. Vanya and Cubes_html/css_WEB-ITnose

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Release: 2016-06-24 11:53:07
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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:








题意:最高层有1个 cube, 第二层有1?+?2?=?3 个cubes, 第三层有1?+?2?+?3?=?6 个 cubes, 依次递增. 然而, 第i层必须有 1?+?2?+?...?+?(i?-?1)?+?i 个 cubes.给一个n,问n个cube最多可以构成多少层。


分析;水题,直接找规律的题。看了一眼,感觉太水,就没敢暴力。。。瞬间想起了以前跟tourist学的一个方法,把个n的区间,都预处理成最高层数,然后查询就是O(1)的了。可惜了,边玩边水,硬是水了一个小时。。。最后看别人的代码,直接暴力也不超时。只能说,数据太弱了。




AC代码:

#include <cstdio>#include <algorithm>using namespace std;int a[10005];int main(){//  freopen("in.txt", "r", stdin);    int foo = 0;    int sum = 0;    for(int i=1; ; i++){        int t = foo;        if(sum > 10000) break;        foo += i;        for(int j=0; j<foo sum j a i-1 foo int n while printf return>  <br>  <br>  <p></p>  <p><br> </p>  <p>切记,下次不能在水题上浪费时间了。。。</p>  <p><br> </p>  <p><br> </p> </foo></algorithm></cstdio>
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