学步园

Pierre is recently obsessed with an online game. To encourage users to log in, this game will give users a continuous login reward. The mechanism of continuous login reward is as follows: If you have not logged in on a certain day, the reward of that day
is 0, otherwise the reward is the previous day's plus 1.

On the other hand, Pierre is very fond of the number N. He wants to get exactly
N points reward with the least possible interruption of continuous login.

Input

There are multiple test cases. The first line of input is an integer T indicates the number of test cases. For each test case:

There is one integer N (1 <= N <= 123456789).

Output

For each test case, output the days of continuous login, separated by a space.

This problem is special judged so any correct answer will be accepted.

Sample Input

4
20
19
6
9

Sample Output

4 4
3 4 2
3
2 3

Hint

20 = (1 + 2 + 3 + 4) + (1 + 2 + 3 + 4)

19 = (1 + 2 + 3) + (1 + 2 + 3 + 4) + (1 + 2)

6 = (1 + 2 + 3)

9 = (1 + 2) + (1 + 2 + 3)

Some problem has a simple, fast and correct solution. 

解题：通过列出组和方式，得知组和的组数最多只有3个，输出原则是：先组数最小，组数相等的每组的数字尽量平分。

首先：用求和分式列出只要一组就能组成的数 。  再查找时，先查只有一组的，如果没有，再查两组的，两组都是建立在一组基础之上。如果两组不能组成就用三组来构成输入的数M。

#include<iostream>
#include<stdio.h>
#include<algorithm>
using namespace std;
#define N 123456789
int ans[20000],n;
int find(int M)
{
    int l,r,m;
    l=1; r=n;
    while(l<=r)
    {
        m=(l+r)/2;
        if(M==ans[m])return m;
        if(M>ans[m])l=m+1;
        else r=m-1;
    }
    if(ans[m]>M)m--;
    return m;
}
int main()
{
    int k[3],M,t,a,b,c,flog;
    for(n=1;n*(n+1)/2<=N;n++)
        ans[n]=n*(n+1)/2;
        n=n-1;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&M);
        k[0]=find(M);
        if(ans[k[0]]==M) printf("%d\n",k[0]);
        else{
            a=M/2; b=M-a; flog=0;
            k[1]=find(b);
            if(ans[k[1]]<b)k[1]++;
            for(;ans[k[1]]<M&&k[1]<=n;k[1]++)
            {
                k[0]=find(M-ans[k[1]]);
                if(ans[k[0]]+ans[k[1]]==M)
                {
                    printf("%d %d\n",k[0],k[1]);
                    flog=1; break;
                }
            }

            if(flog==0)
            {
                int tem;
               a=M/3; k[0]=find(a);
                for(;k[0]>0&&flog==0;k[0]--)
                for(k[1]=k[0];ans[k[0]]+ans[k[1]]<M&&k[1]<=n;k[1]++)
                {
                    c=M-ans[k[0]]-ans[k[1]];
                    k[2]=find(c);
                    if(ans[k[0]]+ans[k[1]]+ans[k[2]]==M)
                    {
                        if(k[2]>k[0]){tem=k[2]; k[2]=k[0]; k[0]=tem; }
                        if(k[2]>k[1]){tem=k[2]; k[2]=k[1]; k[1]=tem; }
                        if(k[0]>k[1]){tem=k[0]; k[0]=k[1]; k[1]=tem; }
                        printf("%d %d %d\n",k[0],k[1],k[2]);
                        flog=1; break;
                    }
                }
            }
        }
    }
}