LightOJ 1045 - Digits of Factorial (k进制下N!的位数)
2016-02-08 17:17
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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) and
base (2 ≤ base ≤ 1000). Both ofthese integers will be given in decimal.
题意:f
=n!,求f
在k进制下的位数。
思路:f
在k进制下的位数,即logk(f
)=log10(f
)/log10[k],打个1e6的表,表示10进制下f
的位数,即log10(f
),然后结果向上取整就行了,当n=0的时候加个特判。。。
ac代码:
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) and
base (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 |
题意:f
=n!,求f
在k进制下的位数。
思路:f
在k进制下的位数,即logk(f
)=log10(f
)/log10[k],打个1e6的表,表示10进制下f
的位数,即log10(f
),然后结果向上取整就行了,当n=0的时候加个特判。。。
ac代码:
#include<stdio.h> #include<math.h> #include<string.h> #include<stack> #include<set> #include<queue> #include<vector> #include<iostream> #include<algorithm> #define MAXN 1010100 #define LL long long #define ll __int64 #define INF 0x7fffffff #define mem(x) memset(x,0,sizeof(x)) #define PI acos(-1) #define eps 1e-10 using namespace std; int gcd(int a,int b){return b?gcd(b,a%b):a;} int lcm(int a,int b){return a/gcd(a,b)*b;} LL powmod(LL a,LL b,LL MOD){LL ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;} //head double a[MAXN]; void debug() { for(int i=1;i<=10;i++) printf("%lf\n",a[i]); } void db() { a[0]=log10(1); for(int i=1;i<=1000000;i++) a[i]=a[i-1]+log10(i); //debug(); } int main() { db(); int t,i,j; int n,k; int cas=0; scanf("%d",&t); while(t--) { scanf("%d%d",&n,&k); printf("Case %d: ",++cas); if(n==0) { printf("1\n"); continue; } double ans; ans=ceil(a /log10(k*1.0)); printf("%d\n",(int)ans); } return 0; }
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