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


[align=left]Problem Description[/align]
iSea is tired of writing the story of Harry Potter, so, lucky you, solving the following problem is enough.



[align=left]Input[/align]
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

[align=left]Output[/align]
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).

[align=left]Sample Input[/align]

2

2 2

10 10

[align=left]Sample Output[/align]

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。

#include<stdio.h>
#include<algorithm>
#include<iostream>
#include<queue>
#include<set>
#include<math.h>
#include<string.h>
using namespace std;
typedef unsigned long long LL;
bool prime[10000015];
LL ans[1000000];
LL prime_table[10000];
LL pr_cnt[10000];
LL pr_cn[10000];
LL slove(LL n,LL m,int cn);
int main(void)
{
int T;
scanf("%d",&T);
int i,j;
for(i = 2; i < 10000; i++)
{
if(!prime[i])
{
for(j = i; (i*j) <= 10000010; j++)
{
prime[i*j] = true;
}
}
}
int cn = 0;
for(i = 2; i < 10000010; i++)
if(!prime[i])ans[cn++] = i;
int __ca = 0;
while(T--)
{
LL n,m;
__ca++;
scanf("%llu %llu",&m,&n);
printf("Case %d: ",__ca);
if(n == 1)
printf("inf\n");
else
{
printf("%llu\n",slove(n,m,cn));
}
}
return 0;
}
LL slove(LL n,LL m,int cn)
{
int bn = 0;
int f = 0;
bool flag  = false ;
memset(pr_cnt,0,sizeof(pr_cnt));
memset(pr_cn,0,sizeof(pr_cn));
while(n>1)
{
while(n%ans[f]==0)
{
if(!flag)
{
flag = true;
bn++;
prime_table[bn] = ans[f];
}
pr_cnt[bn]++;
n/=ans[f];
}
f++;
flag = false;
if((LL)ans[f]*(LL)ans[f] > n)
break;
}
if(n > 1)
{
bn++;
prime_table[bn] = n;
pr_cnt[bn]++;
}//printf("%d\n",n);
LL maxx = -1;
for(int i = 1; i <= bn; i++)
{      //printf("%llu\n",prime_table[i]);
LL v = m;
while(v)
{
v/=(LL)prime_table[i];
pr_cn[i]+=v;
}
if(maxx == -1)
{
maxx = (LL)pr_cn[i]/(LL)pr_cnt[i];
}
else
maxx = min((LL)pr_cn[i]/(LL)pr_cnt[i],maxx);
}
return maxx;
}