light oj1045 - Digits of Factorial (阶乘的位数,,运用对数)
2016-10-25 08:32
239 查看
1045 - Digits of Factorial
Factorial of an integer isdefined 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, thefactorial 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) ofthe factorial of an integer in a certain base.
Each case begins with two integers n (0 ≤ n ≤106) andbase (2 ≤ base ≤ 1000). Both ofthese integers will be given in decimal.
这题真的算的上是创新,,,
乍一看连给予X进制的阶乘都不会求还怎么求位数,,,,题目巧妙在可以跳过中间环节,,,,
联想10进制求位数时用log10(x)+1,,,如果从这个式子联想的话,x 进制的数字的sum位数应该是logx(sum);
这里只需要借助一下换底公式就可以跳过计算x阶乘的环节,,,,,还是需要思维+联想,,,
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
const int N = 1000010;
double f
;
int main()
{
f[0]=0;
for(int i=1;i<=N;i++)
{
f[i]=f[i-1]+log10(i*1.0);
}
int t, ncase=1;
scanf("%d", &t);
while(t--)
{
int n, base;
scanf("%d %d", &n, &base);
printf("Case %d: %d\n",ncase++,int(f
/log10(base*1.0))+1);
}
return 0;
}
![]() ![]() | PDF (English) | Statistics | Forum |
Time Limit: 2 second(s) | Memory Limit: 32 MB |
f(0)= 1
f(n) = f(n - 1) * n, if(n> 0)
So, factorial of 5 is 120. But in different bases, thefactorial 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) ofthe 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) andbase (2 ≤ base ≤ 1000). Both ofthese integers will be given in decimal.
Output
For each case of input you have to print the case number andthe 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: 18488885 Case 5: 1 |
乍一看连给予X进制的阶乘都不会求还怎么求位数,,,,题目巧妙在可以跳过中间环节,,,,
联想10进制求位数时用log10(x)+1,,,如果从这个式子联想的话,x 进制的数字的sum位数应该是logx(sum);
这里只需要借助一下换底公式就可以跳过计算x阶乘的环节,,,,,还是需要思维+联想,,,
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
const int N = 1000010;
double f
;
int main()
{
f[0]=0;
for(int i=1;i<=N;i++)
{
f[i]=f[i-1]+log10(i*1.0);
}
int t, ncase=1;
scanf("%d", &t);
while(t--)
{
int n, base;
scanf("%d %d", &n, &base);
printf("Case %d: %d\n",ncase++,int(f
/log10(base*1.0))+1);
}
return 0;
}
相关文章推荐
- lightoj 1045 - Digits of Factorial (数学-log运用--阶乘进制位数)
- LightOJ1045 Digits of Factorial 求n的阶乘在k进制下的位数
- Light OJ 1045 Digits of Factorial(求位数)
- Light OJ:1045 Digits of Factorial(数学+思维+对数)
- LIGHTOJ 1045 - Digits of Factorial 【阶乘取对数】
- lightoj 1045 - Digits of Factorial 取对数
- light oj --Digits of Factorial (一个数的位数问题以及log的公式应用)
- Light-oj-1045 lightoj1045 - Digits of Factorial (N!不同进制的位数)
- Light OJ 1045 - Digits of Factorial
- LightOJ 1045 Digits of Factorial(大数进制)
- lightoj1045 - Digits of Factorial
- LightOJ 1045 - Digits of Factorial 简单数论
- Light oj 1045 (求某个数的阶乘在x进制下的位数)
- LightOJ 1045 Digits of Factorial
- lightoj 1045 - Digits of Factorial 【数学】
- 【Ligth-oj】-1045 - Digits of Factorial(数论,log,好)
- LightOJ 1045 Digits of Factorial
- lightoj 1045 - Digits of Factorial
- Lightoj1045 Digits of Factorial(数论)
- LightOJ 1045 Digits of Factorial【进制问题】