Problem Statement : Given an integer n, you have to find whether it can be expressed as summation of factorials. For given n, you have to report a solution such that
n = x1! + x2! + ... + xn! (xi < xj for all i < j) Input
Input starts with an integer T (≤ 10000), denoting the number of test cases.
Each case starts with a line containing an integer n (1 ≤ n ≤ 1018). Output
For each case, print the case number and the solution in summation of factorial form. If there is no solution then print 'impossible'. There can be multiple solutions, any valid one will do. See the samples for exact formatting.
MEMORY LIMIT : 32 MB
My Code :
using namespace std;
ll fact[25];
map<ll,int>there;
bool Check_ON(int mask,int pos) //Check if pos th bit (from right) of mask is ON
{
if( (mask & (1<<pos) ) == 0 )return false;
return true;
}
void init()
{
fact[0] = 1;
loop(i,1,19)
fact[i] = fact[i-1]*i;
int maxMASK = (1<<20) - 1;
loop(mask,0,maxMASK)
{
int curmask = mask;
ll curN = 0;
loop(bitpos,0,20-1)
{
if(Check_ON(curmask,bitpos))
{
curN+= fact[bitpos];
}
}
there[curN] = mask;
}
}
int main()
{
init();
//cout<<there.size()<<endl;
int tc,cas = 0;
//write();
sfi(tc);
while(tc--)
{
map<ll,int>::iterator it;
ll N;
sfl(N);
CASE(cas);
it = there.find(N);
if(it==there.end())
{
pf("impossible\n");
}
else
{
int mask = there[N];
int flag = 0;
loop(bitpos,0,20-1)
{
if(Check_ON(mask,bitpos))
{
if(flag)
pf("+%d!",bitpos);
else
pf("%d!",bitpos);
flag = 1;
}
}
pf("\n");
}
cout<<there.size()<<endl;
}
return 0;
}
The size of "there" is only 655360 . How did it consume 32 MB ?
