light oj 1045 Digits of Factorial (k进制数的位数)
题目链接 题意：求n！在k进制表示下有多少位。 思路：答案为[ log(1)+log(2)+...+log(N) ]+1 其中log的底数都是K
由于有多组数据，预处理一个log的前缀和。
1/* ***********************************************
2Author :111qqz
3Created Time :Tue 13 Sep 2016 05:13:15 PM CST
4File Name :code/loj/1045.cpp
5************************************************ */
6#include <cstdio>
7#include <cstring>
8#include <iostream>
9#include <algorithm>
10#include <vector>
11#include <queue>
12#include <set>-digits-of-factorial-k进制数的位
13#include <map>
14#include <string>
15#include <cmath>
16#include <cstdlib>
17#include <ctime>
18#define fst first
19#define sec second
20#define lson l,m,rt<<1
21#define rson m+1,r,rt<<1|1
22#define ms(a,x) memset(a,x,sizeof(a))
23typedef long long LL;
24#define pi pair < int ,int >
25#define MP make_pair
26using namespace std;
27const double eps = 1E-8;
28const int dx4[4]={1,0,0,-1};
29const int dy4[4]={0,-1,1,0};
30const int inf = 0x3f3f3f3f;
31const int N=1E6+7;
32int n;
33int base;
34double sum[N];
35int main()
36{
37 #ifndef ONLINE_JUDGE
38 freopen("code/in.txt","r",stdin);
39 #endif
40 sum[0] = 0 ;
41 for ( int i = 1 ; i < N ; i++) sum[i] = sum[i-1] + log(i);
42 int T;
43 cin>>T;
44 int cas = 0 ;
45 while (T--)
46 {
47 scanf("%d%d",&n,&base);
48 double ans = sum[n]/log(base)+1;
49 printf("Case %d: %d\n",++cas,int(ans));
50 }
51 #ifndef ONLINE_JUDGE
52 fclose(stdin);
53 #endif
54 return 0;
55}