思路:
$x^2=kn+1$
$x^2-1=kn$
$(x+1)(x-1)=kn$
令$x+1=k1n1$,$x-1=k2n2$,其中$k1k2=k$,$n1n2=n$
枚举n约数 >=$\sqrt{n}$的,代入验证
去重
//By SiriusRen #include <set> #include <cstdio> using namespace std; #define int long long int n;set<int>s; signed main(){ scanf("%lld",&n); if(!n){puts("None");return 0;} for(int i=1;i*i<=n;i++)if(n%i==0){ int res=n/i; for(int j=1;j<=n;j+=res)if((j+1)%i==0)s.insert(j); for(int j=res-1;j<=n;j+=res)if((j-1)%i==0)s.insert(j); }for(set<int>::iterator it=s.begin();it!=s.end();++it)printf("%d\n",*it); }
来源:https://www.cnblogs.com/SiriusRen/p/6592632.html