HDU 1124 factorial (factorial suffix 0)

Test instructions: Give a number n, return its factorial result suffix has several 0.

Ideas:

First, the decimal quality factor is decomposed to 2*5=10. The n! mass factor decomposition, then decomposition, which should contain a min (2 number, 5 number) suffix 0.

Why do you say that? For example N=15, then {1 2 3 4 5 6 7 8 9, then you can generate 2 of the number has {2,4,6,8,10,12,14}, can produce 5 of only {5,10,15} , only 2 times 5 in prime numbers to Form 10, because only 2 is an even number of primes!!! Then min (2 numbers, 5 numbers) determines the number of 10 that can be generated, and it determines the number of 0 produced.

Want to decompose [1,n] one by one to take the number of statistics 5? No. Assuming n=15, the N/5 get 3, then [1,15] 3 is a multiple of 5 is {5,10,15}, only a multiple of 5 to produce a factor of 5, then count+=3. Then, these 3 numbers may still be large, and may continue to be removed by 5 (for example, 25 can be removed two times). However {5,10,15} has been replaced by {A-i}, this sequence must be a continuous natural number, because all is less than 5, that must not be able to produce another 5. What if N/5 is greater than 5? Then [1,n] in multiples of 5 will certainly become [1,N/5], but also a factorial form, continue to use the above method to continue to remove all 5, until the result of N/5/5/5 ... is less than 5. The number after each N/5 is the result of count.

1#include <bits/stdc++.h>2 #defineLL Long Long3 using namespacestd;4 intF (unsignedintN)5 {6     if(N <5) {7         return 0;8     }9     returnN/5+ F (N/5);Ten } One  A intMain () - { -     //freopen ("Input.txt", "R", stdin); theUnsignedintT, A; -Cin>>T; -      while(t--) -     { +scanf"%d",&a); -printf"%d\n", F (a)); +     } A  at     return 0; -}

AC Code

HDU 1124 factorial (factorial suffix 0)