| Faculty Dividing Powers |
| Time Limit: 4000ms, Special Time Limit:10000ms, Memory Limit:65536KB |
| Total submit users: 40, Accepted users: 22 |
| Problem 12500 : No special judgement |
| Problem description |
| Fred Faculty and Paul Power love big numbers. Day after day Fred chooses a random integer n and he computes n!. His friend Paul amuses himself by computing several powers of his randomly chosen integer k like k2, k3 and so on. On a hot summer day, Fred and Paul got really, really bored, so they decided to play a joke on their buddy Dave Divider. Fred chooses a random integer n while Paul chooses a random integer k. They want Dave to find the biggest integer i such that ki divides n! without a remainder, otherwise they will throw a cake in Dave's face. Because Dave does not like cakes in his face, he wants you to help him finding that integer i. |
| Input |
| The first line contains the number of test cases t (1 ≤ t ≤ 100). Each of the following t lines contains the two numbers n,k (2 ≤ n ≤ 1018, 2 ≤ k ≤ 1012) separated by one space. |
| Output |
| For each test case, print the maximum integer i on a separate line. |
| Sample Input |
2 5 2 10 10 |
| Sample Output |
3 2 |
| Judge Tips |
| Be careful with overflows in this problem (use 64 bit integers, avoid multiplications which will lead to overflow). |
| Problem Source |
| GCPC 2011 |
题意是说给出n , k 求出最大的 i 使得 n! % k^i == 0 …
假设最简单的情况…k是质数…要求 n! = 1*2*3…*n …易看出在k的倍数里..有1个k..在k的平方的有2个k..在k的立方中有3个k… 那么 n! 中k的个数为 n/k+n/(k^2)+n/(n^3)….及为最大的 i …
拓展一步..若k非质数..但只有一个质因子..如8，9，125 之类的…可以先求出在 n! 中有多少个其质因子,设为x…那么有多少个k..就是 i = x/p…p是指k为起质因数的多少次方..
最终拓展出题目所要求的任意数的情况..k=a1^k1 * a2^k2 * a3^k3…an^kn 其中a1,a2…an为质数..可以算出n!中有多少a1,a2,a3…an…而组成一个k需要k1个a1..k2个a2..kn个an..那么也就是说n(a1)/k1 , n(a2)/k2 …. n(an)/kn…中最小的就是答案…
转自：http://www.migantech.com/blog/codes/
1 #include<iostream> 2 #include<cmath> 3 using namespace std; 4 5 bool p[1000010]; 6 __int64 precord[1000010]; 7 __int64 pcnt=0; 8 void init() 9 { 10 memset(p,false,sizeof(p)); 11 for(int i=2;i<=500004;i++) //素数为false 12 if(p[i]==false) 13 for(int j=i+i;j<=1000009;j+=i) 14 p[j]=true; 15 for(int i=2;i<=1000008;i++) 16 if(p[i]==false) 17 precord[pcnt++]=i; 18 } 19 20 __int64 min(__int64 a,__int64 b) 21 { 22 if(a>b) 23 return b; 24 return a; 25 } 26 27 __int64 krecord[100000]; 28 __int64 krecordcnt[100000]; 29 30 int main() 31 { 32 init(); 33 int t; 34 __int64 n,k; 35 scanf("%d",&t); 36 while(t--) 37 { 38 memset(krecord,0,sizeof(krecord)); 39 memset(krecordcnt,0,sizeof(krecordcnt)); 40 scanf("%I64d%I64d",&n,&k); 41 42 __int64 cnt=0; 43 __int64 temp=k; 44 for(int i=0;i<pcnt;i++) 45 { 46 if(precord[i]>temp) 47 break; 48 if(temp%precord[i]==0) 49 { 50 krecord[++cnt]=precord[i]; 51 while(temp%precord[i]==0) 52 { 53 krecordcnt[cnt]++; 54 temp/=precord[i]; 55 } 56 } 57 } 58 if(temp!=1) 59 { 60 krecord[++cnt]=temp; 61 krecordcnt[cnt]=1; 62 } 63 64 __int64 ans=-1; 65 __int64 pp; 66 __int64 count; 67 for(int i=1;i<=cnt;i++) 68 { 69 pp=krecord[i]; 70 count=0; 71 while(1) 72 { 73 count+=n/pp; 74 if (n/krecord[i]>=pp) 75 pp*=krecord[i]; 76 else 77 break; 78 } 79 count/=krecordcnt[i]; 80 if(ans==-1) 81 ans=count; 82 else 83 ans=min(ans,count); 84 } 85 printf("%I64d\n",ans); 86 } 87 return 0; 88 }