Cut the Cake
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 300 Accepted Submission(s): 135
Problem Description
MMM got a big big big cake, and invited all her M friends to eat the cake together. Surprisingly one of her friends HZ took some (N) strawberries which MMM likes very much to decorate the cake (of course they also eat strawberries, not just for decoration). HZ is in charge of the decoration, and he thinks that it's not a big deal that he put the strawberries on the cake randomly one by one. After that, MMM would cut the cake into M pieces of sector with equal size and shape (the last one came to the party will have no cake to eat), and choose one piece first. MMM wants to know the probability that she can get all N strawberries, can you help her? As the cake is so big, all strawberries on it could be treat as points.
Input
First line is the integer T, which means there are T cases.
For each case, two integers M, N indicate the number of her friends and the number of strawberry.
(2 < M, N <= 20, T <= 400)
For each case, two integers M, N indicate the number of her friends and the number of strawberry.
(2 < M, N <= 20, T <= 400)
Output
As the probability could be very small, you should output the probability in the form of a fraction in lowest terms. For each case, output the probability in a single line. Please see the sample for more details.
Sample Input
2
3 3
3 4
3 3
3 4
Sample Output
1/3
4/27
4/27
Source
Recommend
liuyiding
题目意思很容易看懂。
公式就是 n / (m^(n-1))
这个结果可以积分求得。
枚举两个点，位于两边，就是P(n,2) = n*(n-1)
然后两个点形成的角度范围在 0~1/m之间。剩下n-2个点放的概率就是 x^(n-2)
所以积分 从0~1/m x^(n-2) 对x进行积分。积分结果乘上n*(n-1)
要用高精度。
一种是C++ 完全高精度模板，积攒到最后一场网络赛终于用上了，一用就是两题，哈哈。
但是还是JAVA写起来爽，简单
C++版：
/* ***********************************************
Author :kuangbin
Created Time :2013/9/28 星期六 12:54:45
File Name :2013长春网络赛\1004.cpp
************************************************ */ #pragma comment(linker, "/STACK:1024000000,1024000000")
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
/*
* 完全大数模板
* 输出cin>>a
* 输出a.print();
* 注意这个输入不能自动去掉前导0的，可以先读入到char数组，去掉前导0，再用构造函数。
*/
#define MAXN 9999
#define MAXSIZE 1010
#define DLEN 4 class BigNum
{
private:
int a[]; //可以控制大数的位数
int len;
public:
BigNum(){len=;memset(a,,sizeof(a));} //构造函数
BigNum(const int); //将一个int类型的变量转化成大数
BigNum(const char*); //将一个字符串类型的变量转化为大数
BigNum(const BigNum &); //拷贝构造函数
BigNum &operator=(const BigNum &); //重载赋值运算符，大数之间进行赋值运算
friend istream& operator>>(istream&,BigNum&); //重载输入运算符
friend ostream& operator<<(ostream&,BigNum&); //重载输出运算符 BigNum operator+(const BigNum &)const; //重载加法运算符，两个大数之间的相加运算
BigNum operator-(const BigNum &)const; //重载减法运算符，两个大数之间的相减运算
BigNum operator*(const BigNum &)const; //重载乘法运算符，两个大数之间的相乘运算
BigNum operator/(const int &)const; //重载除法运算符，大数对一个整数进行相除运算 BigNum operator^(const int &)const; //大数的n次方运算
int operator%(const int &)const; //大数对一个int类型的变量进行取模运算
bool operator>(const BigNum &T)const; //大数和另一个大数的大小比较
bool operator>(const int &t)const; //大数和一个int类型的变量的大小比较 void print(); //输出大数
};
BigNum::BigNum(const int b) //将一个int类型的变量转化为大数
{
int c,d=b;
len=;
memset(a,,sizeof(a));
while(d>MAXN)
{
c=d-(d/(MAXN+))*(MAXN+);
d=d/(MAXN+);
a[len++]=c;
}
a[len++]=d;
}
BigNum::BigNum(const char *s) //将一个字符串类型的变量转化为大数
{
int t,k,index,L,i;
memset(a,,sizeof(a));
L=strlen(s);
len=L/DLEN;
if(L%DLEN)len++;
index=;
for(i=L-;i>=;i-=DLEN)
{
t=;
k=i-DLEN+;
if(k<)k=;
for(int j=k;j<=i;j++)
t=t*+s[j]-'';
a[index++]=t;
}
}
BigNum::BigNum(const BigNum &T):len(T.len) //拷贝构造函数
{
int i;
memset(a,,sizeof(a));
for(i=;i<len;i++)
a[i]=T.a[i];
}
BigNum & BigNum::operator=(const BigNum &n) //重载赋值运算符，大数之间赋值运算
{
int i;
len=n.len;
memset(a,,sizeof(a));
for(i=;i<len;i++)
a[i]=n.a[i];
return *this;
}
istream& operator>>(istream &in,BigNum &b)
{
char ch[MAXSIZE*];
int i=-;
in>>ch;
int L=strlen(ch);
int count=,sum=;
for(i=L-;i>=;)
{
sum=;
int t=;
for(int j=;j<&&i>=;j++,i--,t*=)
{
sum+=(ch[i]-'')*t;
}
b.a[count]=sum;
count++;
}
b.len=count++;
return in;
}
ostream& operator<<(ostream& out,BigNum& b) //重载输出运算符
{
int i;
cout<<b.a[b.len-];
for(i=b.len-;i>=;i--)
{
printf("%04d",b.a[i]);
}
return out;
}
BigNum BigNum::operator+(const BigNum &T)const //两个大数之间的相加运算
{
BigNum t(*this);
int i,big;
big=T.len>len?T.len:len;
for(i=;i<big;i++)
{
t.a[i]+=T.a[i];
if(t.a[i]>MAXN)
{
t.a[i+]++;
t.a[i]-=MAXN+;
}
}
if(t.a[big]!=)
t.len=big+;
else t.len=big;
return t;
}
BigNum BigNum::operator-(const BigNum &T)const //两个大数之间的相减运算
{
int i,j,big;
bool flag;
BigNum t1,t2;
if(*this>T)
{
t1=*this;
t2=T;
flag=;
}
else
{
t1=T;
t2=*this;
flag=;
}
big=t1.len;
for(i=;i<big;i++)
{
if(t1.a[i]<t2.a[i])
{
j=i+;
while(t1.a[j]==)
j++;
t1.a[j--]--;
while(j>i)
t1.a[j--]+=MAXN;
t1.a[i]+=MAXN+-t2.a[i];
}
else t1.a[i]-=t2.a[i];
}
t1.len=big;
while(t1.a[len-]== && t1.len>)
{
t1.len--;
big--;
}
if(flag)
t1.a[big-]=-t1.a[big-];
return t1;
}
BigNum BigNum::operator*(const BigNum &T)const //两个大数之间的相乘
{
BigNum ret;
int i,j,up;
int temp,temp1;
for(i=;i<len;i++)
{
up=;
for(j=;j<T.len;j++)
{
temp=a[i]*T.a[j]+ret.a[i+j]+up;
if(temp>MAXN)
{
temp1=temp-temp/(MAXN+)*(MAXN+);
up=temp/(MAXN+);
ret.a[i+j]=temp1;
}
else
{
up=;
ret.a[i+j]=temp;
}
}
if(up!=)
ret.a[i+j]=up;
}
ret.len=i+j;
while(ret.a[ret.len-]== && ret.len>)ret.len--;
return ret;
}
BigNum BigNum::operator/(const int &b)const //大数对一个整数进行相除运算
{
BigNum ret;
int i,down=;
for(i=len-;i>=;i--)
{
ret.a[i]=(a[i]+down*(MAXN+))/b;
down=a[i]+down*(MAXN+)-ret.a[i]*b;
}
ret.len=len;
while(ret.a[ret.len-]== && ret.len>)
ret.len--;
return ret;
}
int BigNum::operator%(const int &b)const //大数对一个 int类型的变量进行取模
{
int i,d=;
for(i=len-;i>=;i--)
d=((d*(MAXN+))%b+a[i])%b;
return d;
}
BigNum BigNum::operator^(const int &n)const //大数的n次方运算
{
BigNum t,ret();
int i;
if(n<)exit(-);
if(n==)return ;
if(n==)return *this;
int m=n;
while(m>)
{
t=*this;
for(i=;(i<<)<=m;i<<=)
t=t*t;
m-=i;
ret=ret*t;
if(m==)ret=ret*(*this);
}
return ret;
}
bool BigNum::operator>(const BigNum &T)const //大数和另一个大数的大小比较
{
int ln;
if(len>T.len)return true;
else if(len==T.len)
{
ln=len-;
while(a[ln]==T.a[ln]&&ln>=)
ln--;
if(ln>= && a[ln]>T.a[ln])
return true;
else
return false;
}
else
return false;
}
bool BigNum::operator>(const int &t)const //大数和一个int类型的变量的大小比较
{
BigNum b(t);
return *this>b;
}
void BigNum::print() //输出大数
{
int i;
printf("%d",a[len-]);
for(i=len-;i>=;i--)
printf("%04d",a[i]);
printf("\n");
} int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int m,n;
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&m,&n);
BigNum tt = ;
for(int i = ;i < n;i++)
tt = tt*m;
int tmp = n;
for(int i = ;i <= n;i++)
{
while( tmp%i == && (tt%i == ) )
{
tmp /= i;
tt = tt/i;
}
}
printf("%d/",tmp);
tt.print(); }
return ;
}
JAVA
import java.io.*;
import java.math.*;
import java.util.*; public class Main {
static BigInteger gcd(BigInteger a,BigInteger b)
{
if(b.equals(BigInteger.ZERO))return a;
else return gcd(b,a.mod(b));
}
public static void main(String arg[])
{
int T;
int n,m;
Scanner cin = new Scanner(System.in);
T = cin.nextInt();
while( T > 0 )
{
m = cin.nextInt();
n = cin.nextInt();
BigInteger tmp1 = BigInteger.valueOf(n);
BigInteger tmp2 = BigInteger.valueOf(m).pow(n-1);
BigInteger tt = gcd(tmp1,tmp2);
tmp1 = tmp1.divide(tt);
tmp2 = tmp2.divide(tt);
System.out.println(tmp1+"/"+tmp2);
T--;
}
}
}