Summer Practice Diary (i)

July 6, 2015

The problem is a math problem, there are two places to pay attention to.

First: The topic asks us to divide a number into successive numbers and requires the number of digits to be the least. By the meaning of the topic can be deduced a formula, the number of n= (A + (a+k-1))/2. Where a represents a continuous number and the smallest of the number, K represents the number of consecutive numbers, the origin of the whole formula is arithmetic progression summation formula derivation.

Second: In the solution, although the formula has been deduced, but in the solution, according to the formula, we take the value of traversing all k to solve a, if the traversal from 2 to N will time out. So it's important to note that you need to open the square root from 2 to 2*n, so you can avoid timeouts.

Code:

#include <cstdio>

#include <iostream>

#include <cmath>

#include <algorithm>

using namespace std;

Main ()

{

int T;

scanf ("%d", &t);

 while (t--)

{

int n,k,flag=0;

scanf ("%d", &n);

 for (int i=2;i<=int (sqrt(n*2)); i++)

{

int temp;

Temp=2*n+i-i*i;

if (temp%i==0&& (temp/i)%2==0&&temp>0)

{

K=i;

flag=1;

break;

}

}

if (flag)

{

printf ("%d =", n);

int a= (2*N+K-K*K)/(2*K);

 for (int i=0;i<k;i++)

{

int ans=a+i;

if (i==k-1)

printf ("%d\n", ans);

Else

printf ("%d +", ans);

}

}

Else

printf ("impossible\n");

}

return 0;

}

Summer Practice Diary (i)