题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5312

Sequence

Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)

Total Submission(s): 1336    Accepted Submission(s): 410

Problem Description

Today, Soda has learned a sequence whose
n-th(n≥1)
item is 3n(n?1)+1.
Now he wants to know if an integer m
can be represented as the sum of some items of that sequence. If possible, what are the minimum items needed?

For example, 22=19+1+1+1=7+7+7+1.

Input

There are multiple test cases. The first line of input contains an integerT(1≤T≤104),
indicating the number of test cases. For each test case:

There‘s a line containing an integer m(1≤m≤109).

Output

For each test case, output
?1
if m
cannot be represented as the sum of some items of that sequence, otherwise output the minimum items needed.

Sample Input

10
1
2
3
4
5
6
7
8
22
10

Sample Output

1
2
3
4
5
6
1
2
4
4

Source

BestCoder 1st Anniversary ($)

题目大意：给出一个序列3*n*（n-1）+1。再输入一个m，求构成给定n所需的最小个数。

（序列中的没一个数能够使用若干次）

解题思路：明白一下N*（N-1）/2为三角形数。性质：随意一个自然数都最多可由三个三角形数表示。 题目给的序列是3*n*（n-1）+1就能够转换为6*n*（n-1）/2+1。对于给定的值m，假如m须要k个数来表示。那么其一组解能够表示为m= 6*（K个三角形数的和）+K；
即随意由k个数组成的解 都有 （m-K）%6==0;那么仅仅要找到最小的k即为所求。

此外。对于序列的通式。当n=1或者n=2的时候，就会没有意义，所以对于1和2的时候须要特殊推断一下。

这是一道三角形数的推导及运用题目。

详见代码 。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>

using namespace std;

int a[100005];

bool check1(int m)
{
    for (int i=1; i<20010; i++)
    {
        if (a[i]==m)
            return 1;
    }
    return 0;
}

bool check2(int m)
{
    int j;
    for (int i=1,j=20010-1; i<20010&&a[i]<m; i++)
    {
        while (a[i]+a[j]>m)
            j--;
        if (a[i]+a[j]==m)
            return 1;
    }
    return 0;
}

int main()
{
    for (int i=0; i<20010; i++)
    {
        a[i]=3*i*(i-1)+1;
    }
    int t;
    scanf("%d",&t);
    while (t--)
    {
        int m;
        scanf("%d",&m);
        int flag=0;
        if (check1(m))
        {
            printf ("1\n");
            continue;
        }
        else if (check2(m))
        {
            printf ("2\n");
            continue;
        }
        else
        {
            for (int i=3; i<=8; i++) //循环到8的原因是由于模6的余数仅仅有0-5六个
            {
                if ((m-i)%6==0)
                {
                    printf ("%d\n",i);
                    flag=1;
                    break;
                }
            }
        }
        if (flag==0)
        {
            printf ("-1\n");

        }
    }
    return 0;
}