Again Prime? No time.
Input: standard input
Output: standard output
Time Limit: 1 second
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
that500 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：求N！%M^k==0；最大的k；
思路：先算M的质因子m，并保存质因子的个数，从N开始递减到N/m，计算每个质因子在N！中出现的次数，取最小。
一开始我的思路是取最大的质因子，然后发现错了，然后想是质因子*个数最大的那个，然后想了下40，100就不对了。所以就把所有的质因子都算了一次。（本来以为会超时。。。结果142ms过。o(︶︿︶)o 唉）
转载请注明出处：寻找&星空の孩子
题目连接：http://acm.hust.edu.cn/vjudge/contest/view.action?cid=77956#problem/E
1 #include<stdio.h> 2 #include<string.h> 3 #define LL long long 4 /* 5 bool prime(int x) 6 { 7 for(int i=2;i*i<=x;i++) 8 { 9 if(!(x%i)) return 0; 10 } 11 return 1; 12 }*/ 13 LL a[10005]; 14 inline void Maxprime(LL x) 15 { 16 LL t=1,k=0; 17 for(LL i=2; i<=x; i++) 18 { 19 while(x%i==0) 20 { 21 t=i; 22 a[i]++; 23 x/=i; 24 } 25 } 26 if(t==1) 27 { 28 a[x]=1; 29 } 30 } 31 inline LL Ssum(LL x) 32 { 33 return (x*(x+1))/2; 34 } 35 LL Sumprime(LL x,LL mod) 36 { 37 LL t=0; 38 while(x) 39 { 40 if(x%mod==0) t++,x/=mod; 41 else return t; 42 } 43 return 0; 44 } 45 46 int main() 47 { 48 LL n,m,T,caseT=1; 49 scanf("%lld",&T); 50 while(caseT<=T) 51 { 52 memset(a,0,sizeof(a)); 53 // memset(b,0,sizeof(b)); 54 scanf("%lld%lld",&m,&n); 55 LL mod; 56 Maxprime(m); 57 /* if(mod!=m) 58 { 59 LL tt=0; 60 for(LL i=2;i<n;i++) 61 { 62 if(tt<i*a[i]) 63 { 64 tt=i*a[i]; 65 mod=i; 66 } 67 } 68 }*/ 69 // printf("mod=%d ",mod); 70 71 LL sum,minsum=9999999999; 72 for(LL j=2; j<=m; j++) 73 { 74 75 if(!a[j]) continue; 76 sum=0; 77 for(LL i=n; i>n/j; i--) 78 { 79 LL tmp=Sumprime(i,j); 80 if(tmp) sum+=Ssum(tmp); 81 } 82 if(minsum>sum/a[j]) minsum=sum/a[j]; 83 } 84 printf("Case %lld: ",caseT++); 85 if(!sum) printf("Impossible to divide "); 86 else printf("%lld ",minsum); 87 } 88 return 0; 89 }