BZOJ1406
题意:给定 n,输出x, n>x>=0, 符合x2%n==1, 没有输出“None”;
思路:
n | (x2-1);
设n=axb;
ab | (x+1)(x-1)
有两种情况
a|(x+1),b|(x-1)或者a|(x-1),b|(x+1);
将n的因数k枚举,1<k<√n;
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<set>
#include<iostream>
using namespace std;
typedef long long ll;
set<ll>Q;
int main()
{
ll n,tmp,a,b;
int count=0;
scanf("%lld",&n);
if(n==1)
{
printf("None\n");
}
for(ll i=1; i*i<n; i++)
{
if(n%i==0)
{
ll a=i,b=n/i;
for(int k=0; (tmp=k*b+1)<n; k++) ///b|(x-1),a|(x+1)
{
if((tmp+1)%a==0)
{
//printf("%d\n",tmp);
Q.insert(tmp);
count++;
}
}
for(int k=1; (tmp=k*b-1)<n; k++) ///b|(x+1),a|(x-1)
{
if((tmp-1)%a==0)
{
//printf("%d\n",tmp);
Q.insert(tmp);
count++;
}
}
}
}
if(count==0)
printf("None\n");
else
{
set<ll>::iterator it;
for(it = Q.begin(); it != Q.end(); it++)
cout<<*it<<endl;
}
return 0;
}