解题报告:HDU_3988 Harry Potter and the Hide Story 大素数分解+勒让德

Harry Potter and the Hide Story

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 2836    Accepted Submission(s): 720


[align=left]Problem Description[/align]
iSea is tired of writing the story of Harry Potter, so, lucky you, solving the following problem is enough.



 

[align=left]Input[/align]
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

 

[align=left]Output[/align]
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).
 

[align=left]Sample Input[/align]

2
2 2
10 10

 

[align=left]Sample Output[/align]

Case 1: 1
Case 2: 2

 

[align=left]Author[/align]
iSea@WHU
 

[align=left]Source[/align]
2011 Multi-University Training Contest 15 - Host by WHU

 

[align=left]Recommend[/align]
lcy
 

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| Note

题意：给定N , K 求最大i 满足 (K^i) | (N!) 

思路：先用Pollard_rho算法得到K的唯一分解式，然后用勒让德定理求得答案

代码：

#include<bits/stdc++.h>

using namespace std;

long long fast_mult(long long x,long long y,long long mod){
x %= mod;
long long res =0;
while(y){
if(y&1){
res = (res+x)%mod;
}x<<=1;
x%=mod;
y>>=1;
}return res;
}

long long qpow(long long x,long long y,const long long& mod){
long long res = 1;
while(y){
if(y&1){
res = fast_mult(res,x,mod);
}x = fast_mult(x,x,mod);
y>>=1;
}return res;
}

inline bool check(long long a,long long n,long long x,long long t){
long long res  = qpow(a,x,n);
long long la = res ;
for(int i=1;i<=t;i++){
res = fast_mult(res,res,n);
if(res==1 && la != 1&& la !=n-1){
return true;
}la = res;
}if(res!=1){
return true;
}return false;
}

inline bool IsPrime(long long n){
if(n==2){
return true;
}if((n&1)==0||n<2)return false;
long long x = n-1;
long long t = 0;
while((x&1)==0){
x>>=1;
t++;
}const int check_time = 20;
for(int i=0;i<check_time;i++){
long long a = rand()%(n-1)+1;
if(check(a,n,x,t)){
return false;
}
}return true;
}

long long e[100];
long long em[100];
int num[100];
int tol ;
long long gcd(long long a ,long long b ){
return b?gcd(b,a%b):a;
}

long long Pollard_rho(long long n,long long c){
long long i = 1, k =2;
long long x = rand()%(n-1)+1;
long long y = x;
while(true){
i++;
x = (fast_mult(x,x,n)+c)%n;
long long d = gcd((y-x+n)%n,n);
if(d>1&&d<n)return d;
if(y==x){
return n;
}if(i==k){
k<<=1;
y=x;
}
}
}

void oper(long long n){
if(n==1)return ;
if(IsPrime(n)){
e[tol++]=n;
return ;
}long long p = n;
while(p>=n)p = Pollard_rho(p,rand()%(n-1)+1);
oper(p);
oper(n/p);
}

inline void work(){
sort(e,e+tol);
int all=0;
memset(num,0,sizeof(num));
em[all]=e[0];
num[all]++;
for(int i=1;i<tol;i++){
if(e[i]==em[all]){
num[all]++;
}else {
em[++all]=e[i];
num[all]++;
}
}tol = all+1;

}

inline long long get(int i,long long n){
long long x = em[i],sum=0;
while(x<=n){
sum+=n/x;
if(x*em[i]/em[i]!=x){
break;
}x*=em[i];
}//printf("%I64d %I64d,sum-->%I64d\n",em[i],n,sum);
return sum/num[i];
}

int main()
{
srand(time(NULL));
int T,t = 0;
scanf("%d",&T);
while(T--){
long long a,b;
scanf("%I64d%I64d",&b,&a);
if(a==1LL){
printf("Case %d: inf\n",++t);
continue;
}
tol = 0;
oper(a);
work();

long long ans = 9e18;
for(int i=0;i<tol;i++){
ans = min(ans,get(i,b));
}printf("Case %d: %I64d\n",++t,ans);
}
}