结论:
n!中因子m的个数k为:
k = n/m+n/(m^2)+n/(m^3)+....
int getnum(int n, int m)
{
int sum = 0;
while(n)
{
sum += n/m;
n /= m;
}
return sum;
}
结论:
n!中因子m的个数k为:
k = n/m+n/(m^2)+n/(m^3)+....
int getnum(int n, int m)
{
int sum = 0;
while(n)
{
sum += n/m;
n /= m;
}
return sum;
}