URAL 1120. Sum of Sequential Numbers （数学）

1120. Sum of Sequential Numbers

Time limit: 1.0 second

Memory limit: 64 MB

There is no involute formulation concerning factitiously activity of SKB Kontur in this problem. Moreover, there is no formulation at all.

Input

There is the only number N, 1 ≤ N ≤ 109.

Output

Your program is to output two positive integers
A and Pseparated with a space such that:

N = A + (A + 1) + … + (A + P − 1).
You are to choose a pair with the maximal possible value of P.

Sample

input	output
14	2 4

Problem Author: Leonid Volkov

Problem Source: USU Open Collegiate Programming Contest October'2001 Junior Session

解析：由N = A + (A + 1) + … + (A + P − 1)可计算出P的最大值为sqrt(2*n),直接枚举即可。

AC代码：

#include <bits/stdc++.h>
using namespace std;

int n;
int ansa, ansp;

int f(int a, int b){
    return ((b - a + 1) % 2) ? ((b - a) / 2 * (a + b - 1) + b) : ((b - a + 1) / 2 * (a + b));
}

int main(){
    int n;
    while(scanf("%d", &n) != EOF){
        int p = sqrt((double)(2 * n));      //P的最大值
        while(p){
            int a = (2*n / p - (p - 1))>>1;     //通过P，计算A
            if(a && f(a, a + p - 1) == n){
                ansa = a;
                ansp = p;
                break;
            }
            p --;
        }
        printf("%d %d\n", ansa, ansp);
    }
    return 0;
}