Hackerearth Divide Number problem solution

In this HackerEarth Divide Number problem solution, we have given integer N, you need to find four integers A, B, C, D, such that they're all factors of N (A|N, B|N, C|N, D|N), and N = A + B + C + D. Your goal is to maximize A x B x C x D.


Hackerearth Divide Number problem solution


HackerEarth Divide Number problem solution.

#include<bits/stdc++.h>

typedef unsigned int uint;
typedef long long ll;
typedef unsigned long long ull;
typedef double lf;
typedef long double llf;
typedef std::pair<int,int> pii;

#define xx first
#define yy second

template<typename T> inline T max(T a,T b){return a>b?a:b;}
template<typename T> inline T min(T a,T b){return a<b?a:b;}
template<typename T> inline T abs(T a){return a>0?a:-a;}
template<typename T> inline bool repr(T &a,T b){return a<b?a=b,1:0;}
template<typename T> inline bool repl(T &a,T b){return a>b?a=b,1:0;}
template<typename T> inline T gcd(T a,T b){T t;if(a<b){while(a){t=a;a=b%a;b=t;}return b;}else{while(b){t=b;b=a%b;a=t;}return a;}}
template<typename T> inline T sqr(T x){return x*x;}
#define mp(a,b) std::make_pair(a,b)
#define pb push_back
#define I inline
#define mset(a,b) memset(a,b,sizeof(a))
#define mcpy(a,b) memcpy(a,b,sizeof(a))

#define fo0(i,n) for(int i=0,i##end=n;i<i##end;i++)
#define fo1(i,n) for(int i=1,i##end=n;i<=i##end;i++)
#define fo(i,a,b) for(int i=a,i##end=b;i<=i##end;i++)
#define fd0(i,n) for(int i=(n)-1;~i;i--)
#define fd1(i,n) for(int i=n;i;i--)
#define fd(i,a,b) for(int i=a,i##end=b;i>=i##end;i--)
#define foe(i,x)for(__typeof((x).end())i=(x).begin();i!=(x).end();++i)
#define fre(i,x)for(__typeof((x).rend())i=(x).rbegin();i!=(x).rend();++i)

struct Cg{I char operator()(){return getchar();}};
struct Cp{I void operator()(char x){putchar(x);}};
#define OP operator
#define RT return *this;
#define UC unsigned char
#define RX x=0;UC t=P();while((t<'0'||t>'9')&&t!='-')t=P();bool f=0;\
if(t=='-')t=P(),f=1;x=t-'0';for(t=P();t>='0'&&t<='9';t=P())x=x*10+t-'0'
#define RL if(t=='.'){lf u=0.1;for(t=P();t>='0'&&t<='9';t=P(),u*=0.1)x+=u*(t-'0');}if(f)x=-x
#define RU x=0;UC t=P();while(t<'0'||t>'9')t=P();x=t-'0';for(t=P();t>='0'&&t<='9';t=P())x=x*10+t-'0'
#define TR *this,x;return x;
I bool IS(char x){return x==10||x==13||x==' ';}template<typename T>struct Fr{T P;I Fr&OP,(int&x)
{RX;if(f)x=-x;RT}I OP int(){int x;TR}I Fr&OP,(ll &x){RX;if(f)x=-x;RT}I OP ll(){ll x;TR}I Fr&OP,(char&x)
{for(x=P();IS(x);x=P());RT}I OP char(){char x;TR}I Fr&OP,(char*x){char t=P();for(;IS(t);t=P());if(~t){for(;!IS
(t)&&~t;t=P())*x++=t;}*x++=0;RT}I Fr&OP,(lf&x){RX;RL;RT}I OP lf(){lf x;TR}I Fr&OP,(llf&x){RX;RL;RT}I OP llf()
{llf x;TR}I Fr&OP,(uint&x){RU;RT}I OP uint(){uint x;TR}I Fr&OP,(ull&x){RU;RT}I OP ull(){ull x;TR}};Fr<Cg>in;
#define WI(S) if(x){if(x<0)P('-'),x=-x;UC s[S],c=0;while(x)s[c++]=x%10+'0',x/=10;while(c--)P(s[c]);}else P('0')
#define WL if(y){lf t=0.5;for(int i=y;i--;)t*=0.1;if(x>=0)x+=t;else x-=t,P('-');*this,(ll)(abs(x));P('.');if(x<0)\
x=-x;while(y--){x*=10;x-=floor(x*0.1)*10;P(((int)x)%10+'0');}}else if(x>=0)*this,(ll)(x+0.5);else *this,(ll)(x-0.5);
#define WU(S) if(x){UC s[S],c=0;while(x)s[c++]=x%10+'0',x/=10;while(c--)P(s[c]);}else P('0')
template<typename T>struct Fw{T P;I Fw&OP,(int x){WI(10);RT}I Fw&OP()(int x){WI(10);RT}I Fw&OP,(uint x){WU(10);RT}
I Fw&OP()(uint x){WU(10);RT}I Fw&OP,(ll x){WI(19);RT}I Fw&OP()(ll x){WI(19);RT}I Fw&OP,(ull x){WU(20);RT}I Fw&OP()
(ull x){WU(20);RT}I Fw&OP,(char x){P(x);RT}I Fw&OP()(char x){P(x);RT}I Fw&OP,(const char*x){while(*x)P(*x++);RT}
I Fw&OP()(const char*x){while(*x)P(*x++);RT}I Fw&OP()(lf x,int y){WL;RT}I Fw&OP()(llf x,int y){WL;RT}};Fw<Cp>out;

const int N=2333;
const lf eps=1e-7;

std::pair<int,lf>s[N],t[N];
pii f[N];
int n,c;

inline void dfs(int x,int l,lf r)
{
if(x==n)
{
if(r<eps)
{
int v=1;
fo0(i,n)v=v/gcd(v,t[i].xx)*t[i].xx;
f[c].yy=v;
f[c].xx=1;
fo0(i,n)f[c].xx*=t[i].xx;
c++;
}
}
else
{
if(r<eps)return;
while(s[l].yy*(n-x)>=r-eps)
{
t[x]=s[l];
if(s[l].yy<=r+eps)dfs(x+1,l,r-s[l].yy);
l++;
}
}
}

int main()
{
fo1(i,N-1)s[i]=mp(i,1./i);
n=4;
dfs(0,1,1);
std::sort(f,f+c);
int u=0;
fo0(i,c)
{
bool flag=0;
fo0(j,u)if(f[i].yy%f[j].yy==0)flag=1;
if(!flag)f[u++]=f[i];
}
c=u;
for(int T=in;T--;)
{
int n=in;
ll ans=-1;
fo0(i,c)if(n%f[i].yy==0)
{
ans=(ll)n*n*n*n/f[i].xx;
break;
}
out,ans,'\n';
}
}


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