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
256
YES
512
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 "); 40 else printf("NO"); 41 } 42 }