P1147 continuous natural number and P1147 continuous Natural Number

P1147 continuous natural number and P1147 continuous Natural Number
Description

For a given natural number M, find all consecutive natural number segments. The sum of all numbers in these consecutive natural number segments is M.

Example: 1998 + 1999 + 2000 + 2001 + 2002 = 10000, so a natural number segment from 1998 to 2002 is a solution of M = 10000.

Input/Output Format Input Format:

A single row containing an integer is given the value of M (10 <= M <= 2,000,000 ).

Output Format:

Each line has two natural numbers. The first and last numbers in a continuous natural number segment that meet the conditions are given. The two numbers are separated by a space, the first of all output rows is listed in ascending order from small to large. For a given input data, there must be at least one solution.

Input and Output sample Input example #1:

combo.in10000

Output sample #1:

combo.out18 142 297 328 388 412 1998 2002

In fact, this question is completely unnecessary for mathematics.

Make a prefix and

Then, the brute-force enumeration can be performed.

Note that you need to add paper-cutting. Otherwise, the time-out will occur.

Always remember:

A miracle of violence !!!!

 

 1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cmath> 5 using namespace std; 6 const int MAXN=10000001; 7 int a[MAXN]; 8 int ans=0; 9 int main()10 {11     int n;12     scanf("%d",&n);13     for(int i=1;i<=n;i++)14         a[i]=a[i-1]+i;15     for(int i=1;i<n;i++)16     {17         for(int j=i;j<n;j++)18         {19             if(a[j]-a[i]==n)20             {21                 printf("%d %d\n",i+1,j);22             }23             if(a[j]-a[i]>n)break;24         }25     }26     return 0;27 }