【Ligth-oj】-1045 - Digits of Factorial(数论,log,好)
2016-08-23 21:17
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1045 - Digits of Factorial
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.
Each case begins with two integers n (0 ≤ n ≤ 106) and base (2 ≤ base ≤ 1000). Both of these integers will be given in decimal.
题意:就是给你一个数n,求出他的阶乘,把阶乘转换成k进制会有多少位。如5!=120,120化为8进制就是170,有3位
题解:把一个十进制数怎么换成k进制呢?就是每次除以k得到的余数就是一位。如:120换成8进制。120%8=0,120/8=15;
15%8=7,15/8=1;
1%8=1,1/8=0;那么就是170了(倒着写)
n!<=k^m,求出最小的m就是换算出k进制有多少位了。
1*2*384*......*n=k^m两边分别取对数得:log1+log2+log3+......+logn=m*logk;
那么m=(log1+log2+log3+......+logn)/(logk)
前面log1+....logn可以先打出来
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, 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 | 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 |
题解:把一个十进制数怎么换成k进制呢?就是每次除以k得到的余数就是一位。如:120换成8进制。120%8=0,120/8=15;
15%8=7,15/8=1;
1%8=1,1/8=0;那么就是170了(倒着写)
n!<=k^m,求出最小的m就是换算出k进制有多少位了。
1*2*384*......*n=k^m两边分别取对数得:log1+log2+log3+......+logn=m*logk;
那么m=(log1+log2+log3+......+logn)/(logk)
前面log1+....logn可以先打出来
AC代码:
<span style="font-size:10px;">#include<cstdio> #include<cstring> #include<cmath> #include<algorithm> using namespace std; #define INF 0x3f3f3f3f #define CLR(a,b) memset(a,b,sizeof(a)) #define LL long long double a[1000010]; void init() { a[1]=0; for(int i=2;i<=1000000;i++) a[i]=a[i-1]+log((double)i); //求出 n!取对数的值 } int main() { int u,ca=1; init(); scanf("%d",&u); while(u--) { int n,k; scanf("%d%d",&n,&k); printf("Case %d: ",ca++); if(n==0) printf("1\n"); else { double ans; ans=a /(log((double)k)); if(ans!=(int)ans) ans=(int)ans+1; printf("%.lf\n",ans); } } return 0; }</span><span style="font-size:18px;"> </span>
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