#include<iostream>
#include<cmath>
using namespace std;
int sum (int a,int b) {return (!((b-a+1)%2))?(b-a+1)/2*(a+b):(b-a)/2*(a+b-1)+b;}
int main () {
int n;
cin>>n;
int a_real,p_real;
int p=(int)(sqrt((double)(n*2)));
do {
int a=(((n*2)/p-(p-1))/2);
if (a&&sum(a,a+p-1)==n) {a_real=a;p_real=p;break;}
p--;
}
while (p);
cout<<a_real<<" "<<p_real<<endl;
return 0;
}

strange solution...
there is a simpler way.

//0.015 sec, 108 КiB
#include <stdio.h>
int main(void){
    unsigned n,p,s;
    scanf("%u",&n);
    for(p=44720;n<(s=(p*(p+1))>>1)||(n-s)%p;p--);
    printf("%u %u\n",1+(n-s)/p,p);
    return 0;
}

EXCELLENT! When I got it))