hdu4488 Faulhaber’s Triangle

Description

The sum of the m ^th  powers of the first n integers
S(n,m)  =  SUM  (  j=  1  to  n)( j ^m)

Can be written as a polynomial of degree m+1 in n:

S(n,m)  =  SUM  (k  =  1  to  m+1)(F(m,k)  *n ^k)

Fo  example:



The coefficients F(m,k) of these formulas form Faulhaber‘s Tr angle:


where  rows m start with 0 (at the top) and columns k go from 1 to m+1

Each  row of Faulhaber‘s Tr angle can be computed from the previous  row by:

a)  The element in  row i and column j ( j>1) is (i/j )*(the element above left); that is:
F(i,j ) =  (i/j )*F(i-1, j-1)
b)  The first element in each  row F(i,1) is chosen so the sum of the elements in the  row is 1

Write a program to find entries in Faulhaber‘s Tr angle as decimal f actions in lowest terms

 

Input

The first line of input contains a single integer  P, (1  <=  P <= 1000), which is the number  of data sets that follow.   Each data set should be processed identically and independently

Each data set consists of a single line of input consisting of three space separated decimal integers The first integer  is the data set number. The second integer  is  row number  m, and the third integer  is the index k within the  row of the entry for  which you are to find F(m,  k), the Faulhaber‘s Triangle entry (0  <=  m  <=  400,  1  <=  k  <=  n+1).

 

Output

For each data set there is a single line of output.  It contains the data set number,  followed by a single space which is then followed by either the value if it is an integer  OR by the numerator  of the entry, a forward slash and the denominator  of the entry.

 

Sample Input

    
    

     
     4
1 4 1
2 4 3
3 86 79
4 400 401
    
    

 

Sample Output

    
    

     
     1 -1/30
2 1/3
3 -22388337
4 1/401
    
    

 

 此题的递归方程已经给出：F(i,j ) =  (i/j )*F(i-1, j-1)，就是在处理成分数时比较麻烦，中间求公约数时出了点问题，一晚上无限纠结啊

 

代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#define N 100000
using namespace std;
__int64 f[405][405][2],MIN,sum1,sum2;//f[i][j][1]表示分子，f[i][j][2]表示分母

__int64 getsMIN(__int64 x,__int64 y)//求最大公约数用来约分，不然__int64会爆
{
    x=(x<0)?-x:x;
    y=(y<0)?-y:y;
    __int64 t=x%y;
    while(t!=0)
    {

        x=y;
        y=t;
        t=x%y;
    }
    return y;
}

int main()
{
    f[0][1][1]=1;f[0][1][2]=1;
    f[1][2][1]=1;f[1][2][2]=2;
    f[1][1][1]=1;f[1][1][2]=2;
    int i,j,P,T,m,k,ans;
    for(i=2;i<=400;i++)
    {
        for(j=i+1,sum1=0,sum2=1;j>1;j--)
        {
            f[i][j][1] = i*f[i-1][j-1][1];
            f[i][j][2] = j*f[i-1][j-1][2];
            MIN=getsMIN(f[i][j][1],f[i][j][2]);
            f[i][j][1] /= MIN;
            f[i][j][2] /= MIN;//约分
  //          cout<<"f["<<i<<"]["<<j<<"][1]: f["<<i<<"]["<<j<<"][2]"<<endl;
  //          cout<<f[i][j][1]<<"/"<<f[i][j][2]<<endl;
            if(sum2%f[i][j][2])
            {
                sum1 = sum1*f[i][j][2]+f[i][j][1]*sum2;
                sum2 *= f[i][j][2];
            }
            else sum1 = sum1 + f[i][j][1]*(sum2/f[i][j][2]);
            MIN=getsMIN(sum1,sum2);
            sum1 /= MIN;
            sum2 /= MIN;
        }
        f[i][1][2] = sum2;
        f[i][1][1] = sum2-sum1;//最后一个只能用1减其余分数的和
        MIN=getsMIN(f[i][1][1],f[i][1][2]);
 //       cout<<"MIN "<<MIN<<endl;
            f[i][1][1] /= MIN;
            f[i][1][2] /= MIN;
 //      cout<<"f["<<i<<"]["<<j<<"][1]: f["<<i<<"]["<<j<<"][2]"<<endl;
  //      cout<<f[i][1][1]<<"/"<<f[i][1][2]<<endl;
    }
//    cout<<f[3][1]<<endl;
    scanf("%d",&P);
    while(P--)
    {
        scanf("%d %d %d",&T,&m,&k);
        if(!f[m][k][1])
            printf("%d 0\n",T);
        else
        {
            if(f[m][k][2]==1)
                printf("%d %I64d\n",T,f[m][k][1]);
            else printf("%d %I64d/%I64d\n",T,f[m][k][1],f[m][k][2]);
        }

    }


    return 0;
}