Initialized to 2, there are only 1 prime factors, and the number of factors is 2.
Then every time the factor of a number is greater than the current factor, this number is the next high composite number, and then this number is multiplied by the previous 10 prime numbers, and then pressed back into the heap. Until we find the top 45. Look at the code in detail.
# include & ltiostream & gt
# include & lt queue & gt
Use namespace std
Structural node
{
int fa[ 10]；
int f；
int num
};
int zs[]={2，3，5，7， 1 1， 13， 17， 19，23，29 }；
Priority queue & ltnode & gtq；;
Boolean operator < (node a, node b)
{
If (a.num! =b.num) returns a.num & gtb.num
Otherwise, return to a.f & gtb.f
}
int ans[50]；
International;
Invalid transaction (node & A)
{
int I；
a . f = 1；
for(I = 0； I< 10; i++)a . f * = a . fa[I]+ 1；
}
Void initialization ()
{
Node temp, new _ one
int i，j；
for(I = 0； I< 10; i++)temp . fa[I]= 0；
temp . fa[0]= 1；
Transaction (temporary);
temp . num = 2；
q . push(temp)；
al = 1；
int now _ f = 0；
And (al < = 46)
{
temp = q . top()；
q . pop()；
if(temp . f & gt； Now _f)
{
now _ f = temp.f
ans[al++]= temp . num；
for(I = 0； I< 10; i++)
{
temp . fa[I]++；
Temporary number * = zs [I];
Transaction (temporary);
q . push(temp)；
temp . fa[I]-；
Temporary number/= zs [I];
}
}
}
}
int main()
{
init()；
int n；
while(scanf("%d "，& ampn)！ =EOF)
{
printf("%d\n "，ans[n])；
}
Returns 0;
}