Lightoj-1045-digits of factorial (using logarithms)

1082-array Queries
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Time Limit:3 second (s) Memory limit:64 MB
Given an array with n elements, indexed from 1 to n. Now you'll be a given some queries in the form I J, your task was to find the minimum value from index I to J.

Input
Input starts with an integer T (≤5), denoting the number of test cases.

The first line of a case was a blank line. The next line contains the integers N (1≤n≤105), Q (1≤q≤50000). The next line contains N space separated integers forming the array. there integers range in [0, 105].

The next Q lines would contain a query which is in the form I J (1≤i≤j≤n).

Output
For each test case, print the case number in a single line. Then for each query, the to print a line containing the minimum value between index I and J.

Sample Input
Output for Sample Input
2

5 3
1 31045-digits of factorial
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Time Limit:2 second (s) Memory limit:32 MB
Factorial of an integer was 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 is 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 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 is begins with the integers n (0≤n≤106) and base (2≤base≤1000). Both of these integers would be a given in decimal.

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

Sample Input
Output for Sample Input
5
5 10
8 10
22 3
1000000 2
0 100
Case 1:3
Case 2:5
Case 3:45
Case 4:18,488,885
Case 5:1
3 5
4 4

1 1
10
1 1
Case 1:
1
3
12
Case 2:
10

Problem-solving ideas: Because the number of times can be used to calculate the logarithm, so f[] directly to the logarithm to save up.

Number of digits = (F[n]/log (base) +1) (PS: +1 is because for example log2 (4) = 2, but his digits are three, so to + 1)

#include <iostream>#include<cstring>#include<cmath>#include<cstdio>using namespacestd;typedef unsignedLong Longull;Const intN =1000010;DoubleF[n];voidinit () {f[0] = log (1);  for(intI=1; i<=n;i++) {F[i]= f[i-1]+log (i*1.0); }    return ;} intMain () {init (); intT,n,Base; scanf ("%d",&T);  for(intt=1; t<=t;t++) {scanf ("%d%d",&n,&Base); printf ("Case %d:%d\n", T, (int) (F[n]/log (Base)+1)); }    return 0;}

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Lightoj-1045-digits of factorial (using logarithms)