Harry Potter and the Hide Story(hdu3988)

Harry Potter and the Hide Story

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2809    Accepted Submission(s): 715

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

 思路：素数分解；

当K = 1的时候肯定输出inf；我们将n分解成素数的乘积，那么我们需要找m!分解后含有这些素数的个数cnt[pi]，那么最高次就是min(cnt[pi]/cnt1[pi]);cnt1[pi]为n中pi的个数。

 1 #include<stdio.h>
 2 #include<algorithm>
 3 #include<iostream>
 4 #include<queue>
 5 #include<set>
 6 #include<math.h>
 7 #include<string.h>
 8 using namespace std;
 9 typedef unsigned long long LL;
10 bool prime[10000015];
11 LL ans[1000000];
12 LL prime_table[10000];
13 LL pr_cnt[10000];
14 LL pr_cn[10000];
15 LL slove(LL n,LL m,int cn);
16 int main(void)
17 {
18         int T;
19         scanf("%d",&T);
20         int i,j;
21         for(i = 2; i < 10000; i++)
22         {
23                 if(!prime[i])
24                 {
25                         for(j = i; (i*j) <= 10000010; j++)
26                         {
27                                 prime[i*j] = true;
28                         }
29                 }
30         }
31         int cn = 0;
32         for(i = 2; i < 10000010; i++)
33                 if(!prime[i])ans[cn++] = i;
34         int __ca = 0;
35         while(T--)
36         {
37                 LL n,m;
38                 __ca++;
39                 scanf("%llu %llu",&m,&n);
40                 printf("Case %d: ",__ca);
41                 if(n == 1)
42                         printf("inf\n");
43                 else
44                 {
45                         printf("%llu\n",slove(n,m,cn));
46                 }
47         }
48         return 0;
49 }
50 LL slove(LL n,LL m,int cn)
51 {
52         int bn = 0;
53         int f = 0;
54         bool flag  = false ;
55         memset(pr_cnt,0,sizeof(pr_cnt));
56         memset(pr_cn,0,sizeof(pr_cn));
57         while(n>1)
58         {
59                 while(n%ans[f]==0)
60                 {
61                         if(!flag)
62                         {
63                                 flag = true;
64                                 bn++;
65                                 prime_table[bn] = ans[f];
66                         }
67                         pr_cnt[bn]++;
68                         n/=ans[f];
69                 }
70                 f++;
71                 flag = false;
72                 if((LL)ans[f]*(LL)ans[f] > n)
73                         break;
74         }
75         if(n > 1)
76         {
77                 bn++;
78                 prime_table[bn] = n;
79                 pr_cnt[bn]++;
80         }//printf("%d\n",n);
81         LL maxx = -1;
82         for(int i = 1; i <= bn; i++)
83         {      //printf("%llu\n",prime_table[i]);
84                 LL v = m;
85                 while(v)
86                 {
87                         v/=(LL)prime_table[i];
88                         pr_cn[i]+=v;
89                 }
90                 if(maxx == -1)
91                 {
92                         maxx = (LL)pr_cn[i]/(LL)pr_cnt[i];
93                 }
94                 else
95                         maxx = min((LL)pr_cn[i]/(LL)pr_cnt[i],maxx);
96         }
97         return maxx;
98 }