uva10780 - Again Prime? No time
Again Prime? No time.
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 and n (0. 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
题目大意：给定m和n，求一个最大正整数k，使得m^k能够被n!整除，变相问n！中有多少个m;
思路：开始想到的是从2-n遍历，把每个数中有多少的m都找出来，后来发现缺陷是m不是质数，例如4*9和6，就找不出有6，实际上应该有两个！正确的解法是分解质因数，然后找出有多少对质因数即可。当然要找的是对于每个质数的最小值.
code：#include <iostream>
#include <cmath>
#include <cstring>
#include <cstdio>
#include <map>
#include <algorithm>
using namespace std;
const int M=0x7fffffff;
const int N=10010;
bool ft[N];
int prime[N],t=0;
void init() //打素数表
{
memset(ft,true,sizeof(ft));
ft[0]=ft[1]=false;
for (int i=2;i<N;i++)
{
if (!ft[i]) continue;
prime[++t]=i;
for (int j=2*i;j<N;j+=i)
ft[j]=false;
}
}
int main()
{
init();
int T;
int m,n;
int i,j;
scanf("%d",&T);
for (int ca=1;ca<=T;ca++)
{
scanf("%d %d",&m,&n);
int ans=M,p=1;
while (m>1)
{
int ct=0;
while (m%prime[p]==0) //分解素数，从前往后找
{
ct++;
m/=prime[p];
}
if (ct==0){p++; continue;}
if (prime[p]>n){ans=-1; break;}
int nt=0;
for (i=prime[p];i<=n;i*=prime[p])
nt+=n/i;
ans=min(ans,nt/ct); //选出最小的素数比，也就是ans个素数对
}
printf("Case %d:\n",ca);
if (ans==-1) cout<<"Impossible to divide"<<endl;
else cout<<ans<<endl;
}
}
)
)
