74th LeetCode Weekly Contest Preimage Size of Factorial Zeroes Function

Let f(x) be the number of zeroes at the end of x!. (Recall that x! = 1 * 2 * 3 * ... * x, and by convention, 0! = 1.)

For example, f(3) = 0 because 3! = 6 has no zeroes at the end, while f(11) = 2 because 11! = 39916800 has 2 zeroes at the end. Given K, find how many non-negative integers x have the property that f(x) = K.

Example 1:
Input: K = 0
Output: 5
Explanation: 0!, 1!, 2!, 3!, and 4! end with K = 0 zeroes.

Example 2:
Input: K = 5
Output: 0
Explanation: There is no x such that x! ends in K = 5 zeroes.

Note:

• K will be an integer in the range [0, 10^9].

或许都知道N！0的个数是怎么算的，但倒过来呢，但打表发现...答案就0和5两种可能

我猜是因为规律是除以5造成的吧...

然后我们二分一下K...

class Solution
{
public:
    int numOfZero(int n){
        int num = 0, i;
        for(i=5; i<=n; i*=5)
        {
            num += n/i;
        }
        return num;
    }
    map<int,int>Mp;
    int preimageSizeFZF(int K){
        int l=K,r=K*10+1;
        while(l<r){
            int mid=l+(r-l)/2;
            if(numOfZero(mid)==K){
                return 5;
            }else if(numOfZero(mid)<K){
                l=mid+1;
            }else{
                r=mid;
            }
        }
         return 0;
    }
};