P1147 consecutive natural numbers and (enumeration, mathematics / half)

Copyright: private individuals do question summed up ~ https://blog.csdn.net/tb_youth/article/details/91548922

https://www.luogu.org/problemnew/show/P1147
descriptive heading
for a given natural number M, to obtain a natural number of all consecutive segments, the successive natural numbers and the sum of the total number of segments M.

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

Input Output Format
Input Format:
contains an integer value of M given in a single line (10≤M≤2,000,000).

Output format:
each row two natural numbers, given in a successive natural numbers satisfying the condition number of a first segment and a last number, separated by a space between the two numbers, a first output for all rows in ascending ascending order, for a given input data, to ensure that there is at least one solution.

Input Output Sample
Input Sample # 1:

10000

Output Sample # 1:

18 142 
297 328 
388 412 
1998 2002

/ *
Way1: data not to be considered directly enumerate + half
way2: Faster certainly is mathematical derivation of
* /
Way1 (O (nlogn)):

#include <iostream>
using namespace std;
typedef long long LL;
int main()
{
    LL m;
    cin>>m;
    for(int i = 0; i < m; i++)
    {
        LL s = i,e  = -1;
        LL low = s+1,high = m-1,mid;
        while(low <= high)
        {
            mid = (low + high) / 2;
            LL temp = (s + mid) * (mid - s + 1);//这里int不够
            if(temp < 2 * m)
            {
                low = mid + 1;
            }
            else if(temp > 2 * m)
            {
                high = mid - 1;
            }
            else
            {
                e = mid;
                break;
            }
        }
        if(e != -1)
        {
            cout<<s<<" "<<e<<endl;
        }
    }
    return 0;
}

way2 (O (n)):
mathematical derivation:
Summation The arithmetic:

sum = (s+e) * (e-s+1) / 2
(s+e) *（e-s+1） = 2 * sum
设 s + e = x1,
   e-s+1 = x2;
   解上述方程组：e = (x1+x2-1)/2, s = (x1-x2+1)/2;
   再由第二个表达式：x1 * x2 = 2*sum
   可知2*sum一定为偶数，那么x1,x2可能有两种情况：
   1.两个都是偶数
   2.一奇一偶
   根据 e = (x1+x2-1)/2, s = (x1-x2+1)/2; 并且要是自然数解，
   那么x1+x2-1与x1-x2+1必定要为偶数，
   so可以排除第一种情况，所以x1,x2必定一奇一偶
   根据x1,x2必定一奇一偶，
   枚举其中一个,根据x1 * x2 = 2*sum得出另一个，
   再根据 e = (x1+x2-1)/2, s = (x1-x2+1)/2就可以求出结果了。
   比如枚举x2，根据x1 * x2 = 2*sum,确定x2的上限：
   up <= sqrt(2*sum),
   后续看代码~

#include <iostream>
#include <math.h>
using namespace std;
typedef long long LL;//其实int就够
int main()
{
    LL m;
    cin>>m;
    LL s,e,x1,x2;
    LL up= sqrt(2*m);
    for(int i = up; i > 1; i--)//逆序枚举x2，因为影响s,e起主导作用的是x1
    {
        x2 = i;
        x1 = 2 * m / x2;
        if(((x2&1) && !(x1&1)) || (!(x2&1) && (x1&1)))
        {
            if(x1 * x2 == 2*m)
            {
                s = (x1 - x2 + 1) >> 1;
                e = (x1 + x2 - 1) >> 1;
                cout<<s<<" "<<e<<endl;
            }
        }
    }
    return 0;
}