hdu4488 Faulhaber’s Triangle(模拟题)
Faulhaber’s Triangle
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 157 Accepted Submission(s): 78
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
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).
1 4 1
2 4 3
3 86 79
4 400 401
2 1/3
3 -22388337
4 1/401
#include<stdio.h> struct nod{
__int64 a,b;
}s[405][405]; __int64 gy(__int64 a,__int64 b)
{
__int64 temp;
if(b==0||a==0)
return 0;
if(a<0)
a=-a;
if(b<0)
b=-b;
if(a<b)
{
temp=a;
a=b;
b=temp;
}
while((temp=a%b))
{
a=b;
b=temp;
}
return b;
}
int Init()
{
int i,j,k,n;
__int64 t1,t2,temp;
s[0][1].a=1;
s[0][1].b=1;
s[1][1].a=1;
s[1][1].b=2;
s[1][2].a=1;
s[1][2].b=2;
for(i=2;i<401;i++)
{
for(j=2;j<=i+1;j++)
{
t1=i*s[i-1][j-1].a;
t2=j*s[i-1][j-1].b;
if((temp=gy(t2,t1)))
{
t1/=temp;
t2/=temp;
}
else
{
t1=0;
t2=1;
}
s[i][j].a=t1;
s[i][j].b=t2;
}
t1=0;
t2=1;
for(j=2;j<=i+1;j++)//2项开始求后面的和
{
t1=t2*s[i][j].a+t1*s[i][j].b;
t2=t2*s[i][j].b;
if((temp=gy(t1,t2)))
{
t1/=temp;
t2/=temp;
}
else
{
t1=0;
t2=1;
}
}
t1=t2-t1;
if((temp=gy(t1,t2)))
{
t1/=temp;
t2/=temp;
}
else
{
t1=0;
t2=1;
}
s[i][1].a=t1;
s[i][1].b=t2;
}
return 1;
}
int main()
{
int i,j,k,n,t,no,x,y;
Init();
scanf("%d",&t);
while(t--)
{
scanf("%d%d%d",&no,&x,&y);
printf("%d ",no);
if(s[x][y].a==0)
printf("0\n");
else if(s[x][y].b==1)
printf("%I64d\n",s[x][y].a);
else printf("%I64d/%I64d\n",s[x][y].a,s[x][y].b);
}
return 0;
}
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