UVA 10780 C - Again Prime? No Time.

Question：

The problem statement is very easy. Given a number n you have to determine the largest power of m,

not necessarily prime, that divides n!.

Input

The input file consists of several test cases. The first line in the file is the number of cases to handle.

The following lines are the cases each of which contains two integers m (1 < m < 5000) and n

(0 < n < 10000). The integers are separated by an space. There will be no invalid cases given and

there are not more that 500 test cases.

Output

For each case in the input, print the case number and result in separate lines. The result is either an

integer if m divides n! or a line ‘Impossible to divide’ (without the quotes). Check the sample input

and output format.

Sample Input

2

2 10

2 100

Sample Output

Case 1:

8

Case 2:

97

题意大意：输入两个数n,m，求最大的k使得m^k为n!的约数。

思路：将m质因数分解，并求出每个质因数的数量，找到n!对应的质因数的个数，其系数比值最小的则为K的值

(http://acm.hust.edu.cn/vjudge/contest/121559#problem/C)

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int INF=0x3f3f3f3f;
int prim[10005],vis[10005],curn=0;
void init()
{
memset(vis,0,sizeof(vis));
for(int i=2;i<=10000;i++)
{
if(!vis[i])
{
prim[curn++]=i;
for(int j=i;j*i<=10000;j++)
vis[i*j]=1;
}
}
}   //构建质数数组；
int main()
{
int n,m,k;

4000
init();
scanf("%d",&k);
for(int i=1;i<=k;i++)
{
scanf("%d%d",&m,&n);
printf("Case %d:\n",i);
int t1,t2,t,sum=INF;
for(int j=0;m!=1;j++)
{
t1=t2=0;
while(m%prim[j]==0)
{
t1++;
m/=prim[j];
}  //找到每个质因数的个数
if(t1!=0)
{
t=n;
while(t!=0)
{
t2+=t/prim[j];
t/=prim[j];
}         //找到n!对应的质因数的个数；
sum=min(t2/t1,sum);
}
}
if(sum==0)
printf("Impossible to divide\n");
else printf("%d\n",sum);
}
return 0;
}

体会：重点求n！中质因数的个数while(n!=0) { t2+=n/prim[j];n/=prim[j];}自己去验证，很奇妙