1340 - Story of Tomisu Ghost

阅读原文时间：2023年07月08日阅读：2

1340 - Story of Tomisu Ghost

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

1000 1000

1000 2

10 8

4 2

Case 1: -1

Case 2: 5227616

Case 3: 2

Case 4: 2

Problem Setter: Shahriar Manzoor

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

思路：很简单，只要将前面的数的阶乘素数分解，然后将每个因子的个数要求末尾0的个数然后答案*power（pi,ans/m）；

其实就是把一个数写成k*d^m;中d的数值；

1 #include
2 #include
3 #include
4 #include
5 #include
6 #include
7 #include
8 #include
9 #include
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 }

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