HDU 2446 Shell Pyramid Binary Search

This morning, I finally got rid of the last question D. My second point is an AK.
I still have the copyright to answer this question ~~ More and more interesting!
Let's talk about how to do it.
I heard that I tried to push the formula myself. It feels good.
First, we can see that

A1 = 1, a2 = 3, a3 = 6. Obviously an = (1 + n) * n/2;
How can we find the Sn ??? Sn = Σ an = (1/2) Σ (n * n + n) = 0.5 * (Σ n * n + Σ n );
In this way, it is converted into the sum of consecutive natural numbers and the sum of consecutive natural numbers.
Continuous natural numbers and (1 + n) * n/2
Sum of squares: n (n + 1) (2n + 1)/2
Sum of cubes: [n (n + 1)/2] ^ 2
Bringing the formula into the ultimate simplification ------> Sn = n (n + 1) (n + 2)/6; a beautiful formula was born

Next we can use binary to search for the answer.
I think my code is still very beautiful ~ Despite vulnerabilities

#include<iostream>#define MAXLayer 1111111using namespace std;__int64 layerSum[MAXLayer];__int64 getSum( __int64 m ){ __int64 sum; sum=m*(m+1)/2;    if( sum%3==0 ){      sum/=3;sum*=(m+2);       }       else sum*=(m+2)/3;    return sum;}int BinarySeachLayer( __int64 &num ){ __int64 l=1,r=3810776,m; __int64 sum; while( m=(l+r)/2,l<r ) {    sum=getSum(m);    if( sum>=num ) r=m;    else l=m+1;  }  num-=getSum(m-1);  return (int)m;}int BinarySeachRow( __int64 &num ){ __int64 l=1,r=3810776,m; __int64 sum; while( m=(l+r)/2,l<r ) {    sum=(1+m)*m/2;    if( sum>=num ) r=m;    else l=m+1;  }  num-=(m-1)*m/2;  return (int)m;}int main(){ __int64 N;int T; scanf( "%d",&T ); while( T-- ) {   scanf("%I64d",&N);    int Layer=BinarySeachLayer( N );    int Row=BinarySeachRow( N );    int Column=N;    printf( "%d %d %d\n",Layer,Row,Column );  }  return 0;}