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

Sequence

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.

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).

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.

10
1
2
3
4
5
6
7
8
22
10

1
2
3
4
5
6
1
2
4
4

#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的原因是因为模6的余数只有0-5六个
            {
                if ((m-i)%6==0)
                {
                    printf ("%d\n",i);
                    flag=1;
                    break;
                }
            }
        }
        if (flag==0)
        {
            printf ("-1\n");

        }
    }
    return 0;
}

hdu 5312 Sequence（数学推导——三角形数）

原文：http://blog.csdn.net/qiqi_skystar/article/details/50237931