中石油训练赛 - Faulhaber’s Triangle(打表)

题目描述

The sum of the m-th powers of the first n integers

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

For example:
S(n, 1) = (1 + . . . + n) = (1/2) ∗ n2+ (1/2) ∗ n
S(n, 2) = (1 + . . . + n2) = (1/3) ∗ n3+ (1/2) ∗ n2+ (1/6) ∗ n
S(n, 3) = (1 + . . . + n3) = (1/4) ∗ n4+ (1/2) ∗ n3) + (1/4) ∗ n2
S(n, 4) = (1 + . . . + n4) = (1/5) ∗ n5+ (1/2) ∗ n4) + (1/3) ∗ n3 − (1/30) ∗ n
The coefficients F(m, k) of these formulas form Faulhaber’s Triangle:
1
1/2 1/2
1/6 1/2 1/3
0 1/4 1/2 1/4
-1/30 0 1/3 1/2 1/5
0 -1/12 0 5/12 1/2 1/6
1/42 0 -1/6 0 1/2 1/2 1/7
where rows m start with 0 (at the top) and columns k go from 1 to m + 1
Each row of Faulhaber’s Triangle can be computed from the previous row by:
a) The element in row i and column j (j > 1) is (i/j) ∗ (theelementaboveleft); 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 Triangle as decimal fractions in lowest terms .

输入

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 ≤ m + 1).

输出

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.

样例输入

样例输出

1 -1/30
2 1/3
3 -22388337
4 1/401题目链接：点击查看

题目大意：给出公式：

当j!=1时， $F(i,j)=(i/j)*F(i-1,j-1)$
当j==1时， $F(i,j)=1-\sum_{j=2}^{i+1}F(i,j)$

初始化F(0,1)=1，求指定位置的值

题目分析：这个题目给了很多无用的信息，但读懂需要打表后写一个函数打完表然后直接查询就行了，为了防止爆范围，我特地用了long long分别储存分子和分母，并写了一个化简函数，每次都除以两个数的gcd来约分，并且将分母上的符号转移到分子上方便输出，直接上代码吧，简单打表：

#include<iostream>
#include<string>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
typedef long long LL;const int N=500;LL a[N][N],b[N][N];//分子,分母 void huajian(LL &a,LL &b)
{LL gcd=__gcd(a,b);a/=gcd;b/=gcd;if(a<0&&b<0)//如果都为负数，全部转正{a=-a;b=-b;}if(a>=0&&b<0)//如果符号在分母上，则转移到分子上{a=-a;b=-b;}
}void init()
{a[0][1]=1;b[0][1]=1;for(int i=1;i<=400;i++){LL suma=0;//suma和sumb储存第2~i+1列的值的总和LL sumb=1;//初始化为0/1（分数形式）for(int j=2;j<=i+1;j++){a[i][j]=i*a[i-1][j-1];b[i][j]=j*b[i-1][j-1];huajian(a[i][j],b[i][j]);LL tempa=suma*b[i][j]+sumb*a[i][j];//根据通分求和化简的LL tempb=sumb*b[i][j];huajian(tempa,tempb);suma=tempa;sumb=tempb;}a[i][1]=sumb-suma;//最后给F(i,1)赋值b[i][1]=sumb;huajian(a[i][1],b[i][1]);}
}int main()
{init();int w;cin>>w;while(w--){int num,x,y;scanf("%d%d%d",&num,&x,&y);printf("%d %lld",num,a[x][y]);if(b[x][y]!=1)printf("/%lld",b[x][y]);printf("\n");}return 0;
}