Digits of Factorial
Time Limit: 2000MS Memory Limit: 32768KByte 64 IO Format:%lld & %llu
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
#include<stdio.h> #include<math.h> double ans, f[1000000] = {0}; void getdigit() { for(int i = 1; i <= 1000000; i++) f[i] = f[i-1] + log(i);//表示 log(1 )+log(2)+。。。+log(i) } int main() { int i, j, n, base, t, cas; scanf("%d", &t); getdigit(); for(cas = 1; cas <= t; cas++) { scanf("%d%d",&n, &base); if(n == 0) printf("Case %d: 1\n", cas);//0在任何的进制里都是0, 故个数就是0 else { ans = f[n] / log(base) + 1;//用到了换底公式 printf("Case %d: %d\n",cas, (int)ans); } } return 0; }