Starship Hakodate-maru

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 751    Accepted Submission(s): 518

Problem Description

The
surveyor starship Hakodate-maru is famous for her two fuel containers
with unbounded capacities. They hold the same type of atomic fuel balls.

There, however, is an inconvenience. The shapes of the fuel
containers #1 and #2 are always cubic and regular tetrahedral
respectively. Both of the fuel containers should be either empty or
filled according to their shapes. Otherwise, the fuel balls become
extremely unstable and may explode in the fuel containers. Thus, the
number of fuel balls for the container #1 should be a cubic number (n^3
for some n = 0, 1, 2, 3, ...) and that for the container #2 should be a
tetrahedral number (n(n+1)(n+2)/6 for some n = 0, 1, 2, 3, ...).

Hakodate-maru
is now at the star base Goryokaku preparing for the next mission to
create a precise and detailed chart of stars and interstellar matters.
Both of the fuel containers are now empty. Commander Parus of Goryokaku
will soon send a message to Captain Future of Hakodate-maru on how many
fuel balls Goryokaku can supply. Captain Future should quickly answer to
Commander Parus on how many fuel balls she requests before her ship
leaves Goryokaku. Of course, Captain Future and her officers want as
many fuel balls as possible.

For example, consider the case
Commander Parus offers 151200 fuel balls. If only the fuel container #1
were available (i.e. if the fuel container #2 were unavailable), at most
148877 fuel balls could be put into the fuel container since 148877 =
53 * 53 * 53 < 151200 < 54 * 54 * 54. If only the fuel container
#2 were available, at most 147440 fuel balls could be put into the fuel
container since 147440 = 95 * 96 * 97 / 6 < 151200 < 96 * 97 * 98 /
6. Using both of the fuel containers #1 and #2, 151200 fuel balls can
be put into the fuel containers since 151200 = 39 * 39 * 39 + 81 * 82 *
83 / 6. In this case, Captain Future‘s answer should be "151200".

Commander
Parus‘s offer cannot be greater than 151200 because of the capacity of
the fuel storages of Goryokaku. Captain Future and her officers know
that well.

You are a fuel engineer assigned to Hakodate-maru.
Your duty today is to help Captain Future with calculating the number of
fuel balls she should request.

Input

The
input is a sequence of at most 1024 positive integers. Each line
contains a single integer. The sequence is followed by a zero, which
indicates the end of data and should not be treated as input. You may
assume that none of the input integers is greater than 151200.

Output

The
output is composed of lines, each containing a single integer. Each
output integer should be the greatest integer that is the sum of a
nonnegative cubic number and a nonnegative tetrahedral number and that
is not greater than the corresponding input number. No other characters
should appear in the output.

Sample Input

100
64
50
20
151200
0

Sample Output

99
64
47
20
151200

Source

Asia 2001, Hakodate (Japan)

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题意:大概就是问 a*a*a+b*(b+1)*(b+2)最接近 n 的那个数是多少?

题解:两层for循环扫一遍。。

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
#include <iostream>
#include <map>
using namespace std;
int n;
int main()
{
    while(scanf("%d",&n)!=EOF,n){
        int ans = 1,ans1 = 1;
        for(int i=1;i<=1000;i++){
            if(i*i*i>n) break;
            ans = i;
        }
        for(int i=1;i<=1000;i++){
            if(i*(i+1)*(i+2)>6*n) break;
            ans1 = i;
        }
        int res = 0;
        for(int i=0;i<=ans+1;i++){
            for(int j=0;j<=ans1+1;j++){
                if(i*i*i+j*(j+1)*(j+2)/6>n) {
                    break;
                }
                res = max(res,i*i*i+j*(j+1)*(j+2)/6);
            }
        }
        printf("%d\n",res);
    }
}