Factorial Problem in Base K

Time Limit: 20 Sec??Memory Limit: 256 MB

題目連接

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3621

Description

How many zeros are there in the end of s! if both s and s! are written in base k which is not necessarily to be 10? For general base, the digit order is 0-9,A-Z,a-z(increasingly), for example F4 in base 46 is actually 694 in base 10,and f4 in base 46 is 1890 in base 10.

Input

There are multiple cases(less than 10000). Each case is a line containing two integers s and k(0 ≤ s < 2^63, 2 ≤ k ≤ 62).

Output

For each case, output a single line containing exactly one integer in base 10 indicating the number of zeros in the end of s!.

?

Sample Input

101 2 12 7 ?

Sample Output

3
1

HINT

題意

給你個在k進制下的數S，然后求S！在K進制下，有多少個末尾0
?

題解：

首先在10進制下，我們是怎么做的？我們先對10進行了質因數分解，分解成了2和5，然后我們就統計s!中，2和5各有多少個，然后取最少的就好了

就這樣，我們先對k進行質因數分解，然后我們取最少個數就好了

代碼：

?

//qscqesze #include <cstdio> #include <cmath> #include <cstring> #include <ctime> #include <iostream> #include <algorithm> #include <set> #include <vector> #include <sstream> #include <queue> #include <typeinfo> #include <fstream> #include <map> #include <stack> typedef long long ll; using namespace std; //freopen("D.in","r",stdin); //freopen("D.out","w",stdout); #define sspeed ios_base::sync_with_stdio(0);cin.tie(0) #define test freopen("test.txt","r",stdin) #define maxn 200001 #define mod 10007 #define eps 1e-9 int Num; char CH[20]; //const int inf=0x7fffffff; //нчоч╢С const int inf=0x3f3f3f3f; inline ll read() {ll x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}return x*f; } inline void P(int x) {Num=0;if(!x){putchar('0');puts("");return;}while(x>0)CH[++Num]=x%10,x/=10;while(Num)putchar(CH[Num--]+48);puts(""); } //**************************************************************************************string s; int n; const int p[18]={2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61}; int a[30]; int main() {while(cin>>s>>n){memset(a,0,sizeof(a));ll tmp=0;ll k=1;for(int i=s.size()-1;i>=0;i--){if(s[i]<='9'&&s[i]>='0')tmp+=(s[i]-'0')*k;else if(s[i]<='Z'&&s[i]>='A')tmp+=(s[i]-'A'+10)*k;else tmp+=(s[i]-'a'+36)*k;k*=n;}for(int i=0;i<18;i++){while(n%p[i]==0&&n>0){n/=p[i];a[i]++;}}ll ans=(1LL<<63)-1;for(int i=0;i<18;i++){ll now=tmp,tot=0;while(now>0){now/=p[i];tot+=now;}if(a[i]>0)ans=min(ans,tot/a[i]);}printf("%lld\n",ans);}}

?

轉載于:https://www.cnblogs.com/qscqesze/p/4539970.html

創作挑戰賽新人創作獎勵來咯，堅持創作打卡瓜分現金大獎

總結

以上是生活随笔為你收集整理的zoj 3621 Factorial Problem in Base K 数论 s!后的0个数的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。