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).
Example
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.
写得有点丑~~~
1 #include<iostream>
2 #include<algorithm>
3 #include<cstdio>
4 #include<math.h>
5 #include<string>
6 using namespace std;
7 typedef long long ll;
8
9 int n;
10 ll cal(ll ans){
11 return ans*(ans+1)/2;
12 }
13
14 int main()
15 { while(~scanf("%d",&n)){
16 int flag=0;
17 for(int i=1;i<=sqrt(2*n)+1;i++){
18 ll sum=n-cal(i);
19 ll l=i,r=sqrt(2*n)+1;
20 ll mid;
21 while(l<r){
22 mid=(l+r)/2;
23 if(cal(mid)==sum){
24 flag=1;
25 break;
26 }
27 if(cal(mid)<sum) l=mid;
28 else r=mid;
29 if(r-l==1){
30 if(cal(r)==sum||cal(l)==sum){
31 flag=1;
32 break;
33 }
34 else break;
35 }
36 }
37 if(flag) break;
38 }
39 if(flag) printf("YES\n");
40 else printf("NO");
41 }
42 }