Description
Input
Output
Sample Input
12
Sample Output
1 5 7 11
Analyse:
x^2=k*n+1 -> x^2-1=kn
(x-1)(x+1)=kn-> (x-1)(x+1)=(k1*n1)*(k2*n2); && k1*k2==k; n1*n2==n;
CODE:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#include<string>
#include<queue>
#include<deque>
#include<stack>
#include<map>
#include<set>
#define INF 0x7fffffff
#define SUP 0x80000000
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
typedef long long LL;
const int N=100007;
int main()
{
LL n;
while(scanf("%lld",&n)==1)
{
LL b;
set<LL> ans;
for(LL i=1;i*i<=n;i++){
if(n%i) continue;
b=n/i;
for(LL j=1;j<n;j+=b){
if((j+1)%i==0)
ans.insert(j);
}
for(LL j=b-1;j<n;j+=b){
if((j-1)%i==0)
ans.insert(j);
}
}
set<LL>::iterator it;
if(!ans.size())
printf("None\n");
else{
for(it=ans.begin();it!=ans.end();it++)
printf("%lld\n",*it);
}
}
return 0;
}