HDU 4488 Faulhaber’s Triangle （数学题）

Faulhaber’s Triangle

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 237    Accepted Submission(s): 126

Problem 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

41 4 12 4 33 86 794 400 401

 

Sample Output

1 -1/302 1/33 -223883374 1/401

 

Source

Greater New York 2012

 

Recommend

liuyiding

 

题意：

题目中有很多没有用的信息，主要内容为：
1.注意第二个图即F(m, k)图，给你m和k，要输出F(m, k)。
2.图中的F(m, k)公式：F(i,j ) = (i/j )*F(i-1, j-1)，( j>1)
3.图中每行的和为1

#include <stdio.h>//#define LL long long//#define LLS "%" "l" "l" "d"#define LL __int64 #define LLS "%" "I" "6" "4" "d"struct node {    LL fz,fm;}FF[405][405];//LL gcd(LL a, LL b) {    //if(b == 0) return a;    //return gcd(b,a%b);//}LL gcd(LL a, LL b) {    LL temp;    if(a < 0) a = -a;    if(b < 0) b = -b;    while(b > 0) {        temp = a;        a = b;        b = temp % b;    }    return a;}void init() {    LL tempi,tempj,sumfz,sumfm,fz,fm,gg;    FF[0][1].fz = 1;    FF[0][1].fm = 1;    for(int i=1; i<=400; i++) {        sumfz = 0;        sumfm = 1;        for(int j=2; j<=i+1; j++) {            if(FF[i-1][j-1].fz == 0) {                FF[i][j].fz = 0;                FF[i][j].fm = 1;            } else {                fz = FF[i-1][j-1].fz;                fm = FF[i-1][j-1].fm;                gg = gcd(i,j);                //if(gg < 0) gg = -gg;                tempi = i / gg;                tempj = j / gg;                gg = gcd(tempj,fz);                //if(gg < 0) gg = -gg;                tempj = tempj / gg;                fz = fz / gg;                gg = gcd(tempi,fm);                //if(gg < 0) gg = -gg;                tempi = tempi / gg;                fm = fm / gg;                FF[i][j].fz = tempi * fz;                FF[i][j].fm = tempj * fm;                //if(FF[i][j].fz == 0)                     //FF[i][j].fm = 1;                sumfz = sumfz * FF[i][j].fm + sumfm * FF[i][j].fz;                sumfm = sumfm * FF[i][j].fm;                gg = gcd(sumfz,sumfm);                //if(gg < 0) gg = -gg;                sumfz = sumfz / gg;                sumfm = sumfm / gg;                if(sumfz == 0)                     sumfm = 1;            }                    }        FF[i][1].fz = sumfm - sumfz;        FF[i][1].fm = sumfm;    }}int main() {    //freopen("in.txt","r",stdin);    int t,m,k,cas;    init();    scanf("%d",&t);    while(t--) {        scanf("%d%d%d",&cas,&m,&k);        printf("%d ",cas);        if(FF[m][k].fm == 1) {            printf(LLS"\n",FF[m][k].fz);        }        else {            printf(LLS"/"LLS"\n",FF[m][k].fz,FF[m][k].fm);        }    }}