/* solution:
Will rapresent the moves as R=right, D=Down....
In any grid of size N, we need N moves of each (R,D) (total 2N moves)
ex for the 2x2 grid we have RRDD,RDRD,RDDR,DRRD,DRDR,DDRR (6 moves)
If we arbitrary choose N of one direction first and fill the rest
with other direction,
the problem becomes a combination problem for N choices out of 2N
so we could use the C(n,k) = n!/(k!(n-k)!) for the answer
where n = 2N and k=N
*/
func Euler015(N int) uint64 {
n, k := (2 * N), N
num := utils.Fact(n)
den := new(big.Int).Mul(utils.Fact(k), utils.Fact(n-k))
res := new(big.Int).Div(num, den)
return res.Uint64()
}
func Euler020() int {
sum := 0
sf := utils.Fact(100).String()
for _, c := range sf {
d, _ := strconv.Atoi(string(c))
sum += d
}
return sum
}
/*solution:
will use a similar analysis of that for problem 30.
we find that we need to consider numbers with max 7 digits as for 8-digit nums max sum of factorials will be with 7 digits.
so max = 9,999,999 at first pas.
but the max num we can obtain by summing factorials on a 7 digit number is 7*9! = 2,540,160
that will be our upper bound
*/
func Euler034() int {
limit := int(7 * utils.Fact(9).Int64())
facts := make([]int, 10)
for i := 0; i < 10; i++ {
facts[i] = int(utils.Fact(i).Int64())
}
sum := 0
for i := 3; i < limit; i++ {
sNum := strconv.Itoa(i)
iSum := 0
for _, c := range sNum {
ic, _ := strconv.Atoi(string(c))
iSum += facts[ic]
}
if i == iSum {
sum += i
}
}
return sum
}