Harry Potter and the Hide Story
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2324 Accepted Submission(s): 569
Problem Description
iSea is tired of writing the story of Harry Potter, so, lucky you, solving the following problem is enough.
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
题意：
给两个数n和k，范围题中给出，然后求最大的 i 使得n！%(k^i)==0。
思路：
若想n!%(k^i)==0，那么k的质因数全部包含在n!中，那么找出k的质因数和n!的质因数，若两质因数相等，该质因数在k中含有b个，在n!中含有a个，那么算出a/b，对于所有不同的质因数算出a/b，那么最小的a/b就是所求的 i （木桶效应）。
代码：
1 #include <cstdio>
2 #include <cstring>
3 #include <algorithm>
4 #include <vector>
5 #include <iostream>
6 using namespace std;
7 #define N 10000005
8
9 __int64 n, m;
10 int p[N], tot;
11 int vis[N];
12
13 void init_p(){ //由于k最大为1<<15，那么初始化素数到1<<8就行了
14 int i, j;
15 tot=0;
16 for(i=2;i<N;i++){
17 if(!vis[i]) p[tot++]=i;
18 for(j=0;j<tot&&p[j]*i<N;j++){
19 vis[p[j]*i]=1;
20 if(i%p[j]==0) break;
21 }
22 }
23 }
24
25 __int64 nn(__int64 a,__int64 b){ //算出n!包含质因数b的个数 ，不懂请看《编程之美》第2.2
26 __int64 ans=0;
27 while(a){
28 a/=b;
29 ans+=a;
30 }
31 return ans;
32 }
33
34 main()
35 {
36 __int64 a, b, MINH;
37 init_p();
38 int i, j, k, kase=1;
39 int t;
40 cin>>t;
41 while(t--){
42 scanf("%I64d %I64d",&n,&m);
43 MINH=-1;
44 if(m==1){
45 printf("Case %d: inf\n",kase++);continue;
46 }
47 for(i=0;i<tot&&p[i]<=m;i++){
48 b=0;
49 if(m%p[i]!=0) continue;
50 while(m%p[i]==0){
51 m/=p[i];
52 b++;
53 }
54 a=nn(n,p[i]);
55 a/=b;
56 if(MINH==-1) MINH=a;
57 else MINH=min(MINH,a);
58 }
59 if(m>1){ //由于咱们需要把m全部包含到n!中，防止m还有剩余，所以有了这个函数
60
61 a=nn(n,m);//printf("1111111111\n");
62 if(MINH==-1) MINH=a;
63 else MINH=min(MINH,a);
64 }
65 printf("Case %d: %I64d\n",kase++,MINH);
66 }
67 }
HDU 3988 n!质因数分解,布布扣,bubuko.com