Zoj 3768

Continuous Login

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Time Limit: 2 Seconds      Memory Limit: 131072 KB      Special Judge

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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

4201969

Sample Output

4 43 4 232 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.

我暴力了些数据，发现答案的个数都没超过3，然后就怀着试一试的想法写了个暴力交了，然后就过了，不知道是数据水了，还是真的可以这样。

#include <stdio.h>#include <string.h>#include <vector>#include <algorithm>#include <string>#include <map>#include <iostream>using namespace std;inline int input(){    int ret=0;    bool isN=0;    char c=getchar();    while(c<'0'||c>'9'){        if(c=='-') isN=1;        c=getchar();    }    while(c>='0'&&c<='9'){        ret=ret*10+c-'0';        c=getchar();    }    return isN?-ret:ret;}int v[20000],cnt=0,a,b,t,n,e;map<int,int>m;map<int,int>to;int main(){    a=1,b=1;    m.clear();    to.clear();    while(a<=123456789){        v[cnt]=a;        m[a]=1;        to[a]=cnt;        cnt++;        b++;        a+=b;    }    t=input();    while(t--){        n=input();        if(m[n]!=0){            printf("%d\n",to[n]+1);            continue;        }        int tag=1;        int e=0;        while(v[e]<n) e++;        for(int i=0;i<=e;i++){            if(m[n-v[i]]!=0){                printf("%d %d\n",i+1,to[n-v[i]]+1);                tag=0;                break;            }        }        if(tag){            for(int i=0;i<=e && tag;i++){                for(int j=e;j>=0 && tag;j--){                    a=n-v[i]-v[j];                    if(m[a]!=0){                        printf("%d %d %d\n",i+1,j+1,to[a]+1);                        tag=0;break;                    }                }            }        }    }}