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  • [HDU3988]Harry Potter and the Hide Story

    Description
    iSea is tired of writing the story of Harry Potter, so, lucky you, solving the following problem is enough.

    image.png

    Input
    The first line contains a single integer T, indicating the number of test cases.
    Each test case contains two integers, N and K.

    Technical Specification

    1. 1 <= T <= 500
    2. 1 <= K <= 1 000 000 000 000 00
    3. 1 <= N <= 1 000 000 000 000 000 000

    Output
    For each test case, output the case number first, then the answer, if the answer is bigger than 9 223 372 036 854 775 807, output “inf” (without quote).

    Sample Input

    2
    2 2
    10 10
    

    Sample Output

    Case 1: 1
    Case 2: 2
    

    这题直接上定理吧……

    勒让德定理:在(n!)的素因子分解式中,素数(p)的指数记为(L_p(n!)),则(L_p(n!)=sumlimits_{k=1}lfloorfrac{n}{p^k} floor)

    证明如下:

    (1,2,3,...,n)都分解为标准形式,记其中(p)的指数为(r)的有(m_r)((rgeqslant 1)),则(L_p(n!)=m_1+2m_2+...=sumlimits_{i=1}im_i)

    分组可得(L_p(n!)=sumlimits_{i=1}m_i+sumlimits_{i=2}m_i+...),令(N_r=sumlimits_{i=r}m_i),则(L_p(n!)=sumlimits_{i=1}N_i)

    易得(N_r=sumlimits_{i=r}m_i)恰好是这(n)个数中能整除(p^r)的数的个数,故(N_r)=(lfloorfrac{n}{p^r} floor),故该定理得证

    /*program from Wolfycz*/
    #include<map>
    #include<cmath>
    #include<cstdio>
    #include<vector>
    #include<cstring>
    #include<iostream>
    #include<algorithm>
    #define Fi first
    #define Se second
    #define ll_inf 1e18
    #define MK make_pair
    #define sqr(x) ((x)*(x))
    #define pii pair<int,int>
    #define int_inf 0x7f7f7f7f
    using namespace std;
    typedef long long ll;
    typedef unsigned int ui;
    typedef unsigned long long ull;
    inline char gc(){
    	static char buf[1000000],*p1=buf,*p2=buf;
    	return p1==p2&&(p2=(p1=buf)+fread(buf,1,1000000,stdin),p1==p2)?EOF:*p1++;
    }
    template<typename T>inline T frd(T x){
    	int f=1; char ch=gc();
    	for (;ch<'0'||ch>'9';ch=gc())	if (ch=='-')    f=-1;
    	for (;ch>='0'&&ch<='9';ch=gc())	x=(x<<1)+(x<<3)+ch-'0';
    	return x*f;
    }
    template<typename T>inline T read(T x){
    	int f=1; char ch=getchar();
    	for (;ch<'0'||ch>'9';ch=getchar())	if (ch=='-')	f=-1;
    	for (;ch>='0'&&ch<='9';ch=getchar())	x=(x<<1)+(x<<3)+ch-'0';
    	return x*f;
    }
    inline void print(int x){
    	if (x<0)	putchar('-'),x=-x;
    	if (x>9)	print(x/10);
    	putchar(x%10+'0');
    }
    const int N=1e7;
    int prime[N+10],tot;
    bool inprime[N+10];
    void prepare(){
    	for (int i=2;i<=N;i++){
    		if (!inprime[i])	prime[++tot]=i;
    		for (int j=1;j<=tot&&i*prime[j]<=N;j++){
    			inprime[i*prime[j]]=1;
    			if (i%prime[j]==0)	break;
    		}
    	}
    }
    int main(){
    //	freopen(".in","r",stdin);
    //	freopen(".out","w",stdout);
    	prepare();
    	int T=read(0);
    	for (int Case=1;Case<=T;Case++){
    		ll n=read(0ll),k=read(0ll),Ans=ll_inf;
    		printf("Case %d: ",Case);
    		if (k==1){
    			printf("inf
    ");
    			continue;
    		}
    		for (int i=1;i<=tot;i++){
    			if (k%prime[i])	continue;
    			ll cntk=0,cntn=0,_n=n;
    			while (k%prime[i]==0)	k/=prime[i],cntk++;
    			while (_n/=prime[i])	cntn+=_n;
    			Ans=min(Ans,cntn/cntk);
    			if (k==1)	break;
    		}
    		if (k>1){
    			ll _n=n,cntn=0;
    			while (_n/=k)	cntn+=_n;
    			Ans=min(Ans,cntn);
    		}
    		printf("%lld
    ",Ans);
    	}
    	return 0;
    }
    
    作者:Wolfycz
    本文版权归作者和博客园共有,欢迎转载,但必须在文章开头注明原文出处,否则保留追究法律责任的权利
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  • 原文地址:https://www.cnblogs.com/Wolfycz/p/14932307.html
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