Ultraviolet A 12683 odd and even zeroes

Time Limit: 1000 MS In mathematics, the factorial of a positive integer number N is written as n! And is de ned as follows:
N! = 1 2 3 4: (N 1) n =
Limit n
I = 1
I
The value of 0! Is considered as 1. n! Grows very rapidly with the increase of N. Some values of N!
Are:
0! = 1
1! = 1
2! = 2
3! = 6
4! = 24
5! = 120
10! = 3628800
14! = 87178291200
18! = 6402373705728000
22! = 1124000727777607680000
You can see that for some values of N, N! Has odd number of Trailing zeroes (eg 5 !, 18 !) And for some
Values of N, N! Has even number of Trailing zeroes (eg 0 !, 10 !, 22 !). Given the value of N, your job is
Nd how values of the values 0 !; 1 !; 2 !; 3 !; :; (N 1 )!; N! Has even number of Trailing zeroes.
Input
Input le contains at most 1000 lines of input. Each line contains an integer N (0 N 10
18
). Input
Is terminated by a line containing a '-1 '.
Output
For each line of input produce one line of output. This line contains an integer which denotes how
Percent of the numbers 0 !; 1 !; 2 !; 3 !; :; N !, Contains even number of Trailing zeroes.
Sample Input
2
3
10
100
1000
2000
3000
10000
100000
200000
-1
Sample output
3
4
6
61
525
1050
1551
5050
50250

100126

Question: given a number of N, 0! , 1! , 2 !, ..., N! The number of the (n + 1) factorial ending 0 is an even number. (0 <= n <= 10 ^ 18)

Idea: I! The number of 0 at the end is determined by the number of 5 in the factorial. We take 5 numbers as a whole.

1 (5) 1 (10) 1 (15) 1 (20) 2 (25) 1 (30) 1 (35) 1 (40) 1 (45) 2 (50) 1 (55) 1 (60) 1 (65) 1 (70) 2 (75) 1 (80) 1 (85) 1 (90) 1 (95) 2 (100) 1 (105) 1 (110)

1 (115) 1 (120) 3 (125) the former number represents the power of 5 in the number, and the latter represents the number.

We will group the number before 625 as follows:

1 1 1 1 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 1 1 1 3

1 1 1 1 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 1 1 1 3

1 1 1 1 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 1 1 1 3

1 1 1 1 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 1 1 1 3

1 1 1 1 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 1 1 1 1 1 4

So which segments meet the conditions where the number of 0 at the end is an even number? We represent the segment of the first line, (y) is yes, (n) is no

(Y) 1 (n) 1 (y) 1 (n) 1 (y) 2 (y) 1 (n) 1 (y) 1 (n) 1 (y) 2 (y) 1 (n) 1 (y) 1 (n) 1 (y) 2 (y) 1 (n) 1 (y) 1 (n) 1 (y) 2 (y) 1 (n) 1 (y) 1 (n) 1 (y) 3

We found that:

(1) It is known that until the first k power of 5 appears, then until the first 5 (k + 1) appears) the power-wise section must be a K-power section of 5 that repeats 5 segments, and the last segment of the last segment

An element changes K to (k + 1), that is, a large segment that appears until the first 5 (k + 1) power is reached.

(2) It is known that the Y/N values of each small segment in the first k POWER segment of the first 5 can be inferred from the (k + 1) of the first 5) y/N of each small segment of the power.

1. If K is an even number, the next four segments are exactly the same as the previous segments.

2. If K is an odd number, section 2nd and section 4th in the next 4 are the same as those in the previous section. Paragraphs 1st and 3rd are the opposite of the previous ones.

If DP [I] [0] is set, it indicates the number of segments whose last 0 is an even number until the first 5 I power is displayed.

DP [I] [1] is the number of even numbers opposite to the above case

So

DP [I] [0] = 5 * DP [I-1] [0] I is an odd number;

DP [I] [0] = 3 * DP [I] [0] + 2 * DP [I] [1] I is an even number;

DP [I] [1] = A [I-1]-DP [I] [0];

After the DP array is pre-processed, for N, we can find the maximum power range not greater than N for every binary query. We may set it to K, x = N/A [K], then N-= A [k] * X.

Occasionally, update ans. At the same time, we set a variable now to record the current state.

# Include <iostream> # include <algorithm> # include <cstdio> # define ll long longusing namespace STD; const int maxn = 27; ll n, a [maxn], DP [maxn] [2]; void initial () {ll T, sum = 1; A [0] = 1; for (INT I = 1; I <maxn; I ++) A [I] = 5 * A [I-1]; DP [0] [0] = 1, DP [0] [1] = 0; // This assignment is intended for later calculation, not 5 times. The values below are 5 times. DP [1] [0] = 1, DP [1] [1] = 0; For (INT I = 2; I <maxn; I ++) {if (I % 2 = 0) DP [I] [0] = 3 * DP [I-1] [0] + 2 * DP [I-1] [1]; else DP [I] [0] = 5 * DP [I-1] [0]; DP [I] [1] = A [I-1]-DP [I] [0];} For (INT I = 1; I <maxn; I ++) {DP [I] [0] * = 5; DP [I] [1] * = 5 ;}} void solve () {ll ans = 0; if (n <= 4) ans = n + 1; else {bool now = 0; n ++; while (n) {int T = upper_bound (A, A + maxn, n)-A-1; ll num = N/A [T]; n = n % A [T]; If (t = 0) ans + = num * DP [T] [now]; else if (T % 2 = 1) ans = ans + (Num + 1) /2 * DP [T] [now] + num/2 * DP [T] [now ^ 1]; else ans = ans + num * DP [T] [now]; N = n % A [T]; If (T % 2 = 1 & num % 2 = 1) Now ^ = 1 ;}} cout <ans <Endl;} int main () {initial (); While (CIN> N) {If (n =-1) break; solve ();} return 0 ;}

Ultraviolet A 12683 odd and even zeroes