Lightoj1045 Digits of Factorial(数论)
2016-08-23 20:24
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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
把n!(十进制)转化为base进制有多少位
取位数用log 但是直接取会爆 可以把它拆开
log(n!) = log(n)+log(n-1) + log(n-2) +...+log(2)
#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
#include<algorithm>
using namespace std;
double dp[1000001];
int main()
{
int t,n,base;
scanf("%d",&t);
dp[0]=0;
for(int i=1;i<1000001;i++)
{
dp[i]+=dp[i-1]+log(i);
}
for(int i=1;i<=t;i++)
{
scanf("%d%d",&n,&base);
printf("Case %d: %d\n",i,int(dp
/log(base))+1);
}
return 0;
}
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
把n!(十进制)转化为base进制有多少位
取位数用log 但是直接取会爆 可以把它拆开
log(n!) = log(n)+log(n-1) + log(n-2) +...+log(2)
#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
#include<algorithm>
using namespace std;
double dp[1000001];
int main()
{
int t,n,base;
scanf("%d",&t);
dp[0]=0;
for(int i=1;i<1000001;i++)
{
dp[i]+=dp[i-1]+log(i);
}
for(int i=1;i<=t;i++)
{
scanf("%d%d",&n,&base);
printf("Case %d: %d\n",i,int(dp
/log(base))+1);
}
return 0;
}
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