As you very well know, this year's funkiest numbers are so called triangular numbers (that is, integers that are representable as , where k is some positive integer), and the coolest numbers are those that are representable as a sum of two triangular numbers.
A well-known hipster Andrew adores everything funky and cool but unfortunately, he isn't good at maths. Given number n, help him define whether this number can be represented by a sum of two triangular numbers (not necessarily different)!
Input
The first input line contains an integer n (1 ≤ n ≤ 109).
Output
Print "YES" (without the quotes), if n can be represented as a sum of two triangular numbers, otherwise print "NO" (without the quotes).
Examples
Input
256
Output
YES
Input
512
Output
NO
Note
In the first sample number .
In the second sample number 512 can not be represented as a sum of two triangular numbers.
这个题我先打了一个表，然后遍历第一个，再用二分找第二个。
#include <iostream>
#include <string>
#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <vector>
#include <cmath>
#include <list>
#include <queue>
#include <stack>
#include <map>
#include <set>
using namespace std;
typedef long long ll;
const ll maxn=1e9+10;
ll f[1000000];
int main()
{ ll s,d;
s=d=1;
int flag=0;
ll cnt=1;
while(s<maxn)
{
f[cnt++]=s;
// printf("%d ",f[cnt-1]);
d=s;
s=d+cnt;
}
ll n;
cin>>n;
for(ll i=1;i<cnt;i++)
{
ll l=i,r=cnt,mid=(l+r)/2;
ll A=f[l],B;
while(l<=r)
{
B=f[mid];
if(A+B<n)
{
l=mid+1;
mid=(l+r)/2;
}
else if(A+B>n)
{
r=mid-1;
mid=(l+r)/2;
}
else {
flag=1;
printf("YES\n");
break;
}
}
if(flag==1) break;
}
if(flag==0) printf("NO\n");
return 0;
}