http://acm.hdu.edu.cn/showproblem.php?pid=1023
//
C(0) = 1 ; (n+2)*C(n+1) = (4n+2)*C(n);
#include
<
iostream
>
using
namespace
std;
int
a[
101
][
101
]
=
{0}
;
int
main()
{
int n,i,j,len,r,temp,t;
int b[101];
a[1][0] = 1;
len = 1;
b[1] = 1;
for(i=2;i<=100;i++)
{
t = i-1;
for(j=0;j<len;j++) //乘法
a[i][j] = a[i-1][j]*(4*t+2);
for(r=j=0;j<len;j++) //处理相乘结果
{
temp = a[i][j] + r;
a[i][j] = temp % 10;
r = temp / 10;
}
while(r) //进位处理
{
a[i][len++] = r % 10;
r /= 10;
}

for(j=len-1,r=0;j>=0;j--) //除法
{
temp = r*10 + a[i][j];
a[i][j] = temp/(t+2);
r = temp%(t+2);
}
while(!a[i][len-1]) //高位零处理
len --;
b[i] = len;
}
while(cin>>n)
{
for(j=b[n]-1;j>=0;j--)
printf("%d",a[n][j]);
printf("/n");
}
return 0;
}













































