Light 1045 Digits of Factorial 【数论】

题目链接：http://acm.hust.edu.cn/vjudge/contest/view.action?cid=120197#problem/U

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Digits of Factorial
Time Limit:2000MS Memory Limit:32768KB 64bit IO Format:%lld & %llu
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Status
Description
Factorial of an integer is defined by the following function

f(0) = 1
f(n) = f(n - 1) * n, if(n > 0)

So, factorial of 5 is 120. But in different bases, the factorial may be different. For example, factorial of 5 in base 8 is 170.

In this problem, you have to find the number of digit(s) of the factorial of an integer in a certain base.

Input
Input starts with an integer T (≤ 50000), denoting the number of test cases.

Each case begins with two integers n (0 ≤ n ≤ 106) and base (2 ≤ base ≤ 1000). Both of these integers will be given in decimal.

Output
For each case of input you have to print the case number and the digit(s) of factorial n in the given base.

Sample Input
5
5 10
8 10
22 3
1000000 2
0 100
Sample Output
Case 1: 3
Case 2: 5
Case 3: 45
Case 4: 18488885
Case 5: 1

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题目大意　　：　　就是求n!在base进制下表示 不算前导0的话 有多少位

多少位的话 直接取log即可
但是n大了 long long 会爆 所以
拆成
log(n!)=log(n)+log(n-1)+ …. （以上底数为base）.

因为计算机中只有底数为十的对数函数 所以 表示成lg(n!)/lg(base) 就行

最后n==0 的时候要特判一下.

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上面就是题解了
但是 我做的时候出现了 迷之RE+迷之WA

RE ： printf(“Case %d: %.0llf\n”,++p,f[n]/log(base)+0.5);
这么输出居然是RE

于是又换了一种

WA：printf(“Case %d: %d\n”,++p,(int)(f[n]/log(1.0*base)+0.5));

这样居然是WA ？我就不明白了

最后还是看别人代码 用的ceil() 才过的 但这到底是为什么啊啊啊啊啊啊啊啊啊
最终过得姿势 ::: printf(“Case %d: %d\n”,++p,(int)ceil(f[n]/log(1.0*base)))

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附本题代码

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#include <stdio.h>#include <stdlib.h>#include <string.h>#include <limits.h>#include <malloc.h>#include <ctype.h>#include <math.h>#include <string>#include <iostream>#include <algorithm>#include <map>#include <vector>#include <set>using namespace std;#define LL long long intdouble f[1001110];int main(){    f[0]=0;    for(int i=1;i<=1000010;i++)    {        f[i]=f[i-1]+log(1.0*i);    }    int t,p=0,n,base;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&n,&base);        if(n==0)        {            printf("Case %d: 1\n",++p);            continue;        }        printf("Case %d: %d\n",++p,(int)ceil(f[n]/log(1.0*base)));    }    return 0;}