1340 - Story of Tomisu Ghost

1340 - Story of Tomisu Ghost

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Time Limit: 2 second(s)

Memory Limit: 32 MB

It is now 2150 AD and problem-setters are having a horrified time as the ghost of a problem-setter from the past, Mr. Tomisu, is frequently disturbing them. As always is the case in most common ghost stories, Mr. Tomisu has an unfulfilled dream: he had set 999 problems throughout his whole life but never had the leisure to set the 1000th problem. Being a ghost he cannot set problems now so he randomly asks problem-setters to complete one of his unfinished problems. One problem-setter tried to convince him saying that he should not regret as 999 is nowhere near 1024 (2¹⁰) and he should not worry about power of 10 being an IT ghost. But the ghost slapped him hard after hearing this. So at last one problem setter decides to complete his problem:

"n! (factorial n) has at least t trailing zeroes in b based number system. Given the value of n and t, what is the maximum possible value of b?"

Input

Input starts with an integer T (≤ 4000), denoting the number of test cases.

Each case contains two integers n (1 < n ≤ 10⁵) and t (0 < t ≤ 1000). Both n and t will be given in decimal (base 10).

Output

For each case, print the case number and the maximum possible value of b. Since b can be very large, so print b modulo 10000019. If such a base cannot be found then print -1 instead.

Sample Input	Output for Sample Input
4 1000 1000 1000 2 10 8 4 2	Case 1: -1 Case 2: 5227616 Case 3: 2 Case 4: 2

 

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Problem Setter: Shahriar Manzoor

Special Thanks: Jane Alam Jan, Md. Towhidul Islam Talukder

思路：很簡單，只要將前面的數的階乘素數分解，而後將每一個因子的個數要求末尾0的個數而後答案*power（pi,ans/m）；

其實就是把一個數寫成k*d^m;中d的數值；

 1 #include <cstdio>
 2 #include <cstdlib>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <iostream>
 6 #include <algorithm>
 7 #include <map>
 8 #include <queue>
 9 #include <vector>
10 using namespace std;
11 typedef  long long LL;
12 const int N=10000019;
13 bool prime[100005];
14 int ans_prime[100005];
15 int fen_prime[100005];
16 LL quick(LL n,LL m);
17 int main(void)
18 {
19         int i,j,k;
20         int an=0;
21         for(i=2; i<=100000; i++)
22         {
23                 if(!prime[i])
24                 {
25                         for(j=i; ((LL)i*(LL)j)<=100000; j++)
26                         {
27                                 prime[i*j]=true;
28                         }
29                         ans_prime[an++]=i;
30                 }
31         }
32         int __ca=0;
33         scanf("%d",&k);
34         while(k--)
35         {
36                 __ca++;
37                 LL n,m;
38                 scanf("%lld %lld",&n,&m);
39                 memset(fen_prime,0,sizeof(fen_prime));
40                 for(i=0; i<an; i++)
41                 {
42                         LL ask=n;
43                         if(ask<ans_prime[i])
44                         {
45                                 break;
46                         }
47                         else
48                         {
49                                 while(ask)
50                                 {
51                                         fen_prime[i]+=ask/ans_prime[i];
52                                         ask/=ans_prime[i];
53                                 }
54                         }
55                         if(fen_prime[i]<m)
56                             break;
57                 }
58                 LL anw=1;
59                 for(i=0;i<an;i++)
60                 {
61                     if(fen_prime[i]<m)
62                     {
63                         break;
64                     }
65                     else
66                     {
67                         anw=(anw*quick(ans_prime[i],fen_prime[i]/m))%N;
68                     }
69                 }
70                 printf("Case %d: ",__ca);
71                 if(anw==1)
72                 {
73                     printf("-1\n");
74                 }
75                 else
76                 {
77                     printf("%lld\n",anw);
78                 }
79         }
80         return 0;
81 }
82 LL quick(LL n,LL m)
83 {
84         LL ak=1;
85         while(m)
86         {
87                 if(m&1)
88                 {
89                         ak=ak*n%N;
90                 }
91                 n=n*n%N;
92                 m/=2;
93         }
94         return ak;
95 }