ZOJ3768 Continuous Login

Continuous Login

---

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

---

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.

---

题目的意思是给出一个数n是的n分成若干的数的和，每个数是重1累加奥x的，保证数字个数最少

找规律发现最多出现3项，分别对1项2项3项讨论，枚举加二分

#include <iostream>#include<queue>#include<cstdio>#include<algorithm>#include<cmath>#include<set>#include <cstring>using namespace std;#define LL long longint a[100005];int main(){    int T,n; a[0]=0;   for(int i=1;i<100005;i++)    a[i]=a[i-1]+i;    scanf("%d",&T);    while(T--)    {        int flag=0;        scanf("%d",&n);       for(int i=0;i<100000;i++)       {           if(a[i]==n){               flag=1;               printf("%d\n",i);               break;           }           if(a[i]>n) break;       }        if(flag) continue;       for(int i=0;a[i]<n;i++)       {           int x=n-a[i];           int pos=lower_bound(a,a+100000,x)-a;           if(a[pos]==x)           {               flag=1;               printf("%d %d\n",i,pos);               break;           }       }       if(flag) continue;       for(int i=0;a[i]<n;i++)       {           for(int j=0;a[j]<n-a[i];j++)           {              int x=n-a[i]-a[j];           int pos=lower_bound(a,a+100000,x)-a;           if(a[pos]==x)           {               flag=1;               printf("%d %d %d\n",i,j,pos);               break;           }           }           if(flag) break;       }    }    return 0;}