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Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7234 Accepted Submission(s): 3896
#include<stdio.h>
#include<string.h>
int a[105][105];
void C()
{
int i,j,k,m,l;
a[1][0]=1;
a[1][1]=1;
a[2][0]=1;
a[2][1]=2;
l=1;
for(j=3;j<=100;j++)
{
k=0;
for(i=1;i<=l;i++)
{
m=a[j-1][i]*(4*j-2)+k;
a[j][i]=m%10;
k=m/10;
}
while(k)
{
a[j][++l]=k%10;
k=k/10;
}
for(i=l;i>=1;i--)
{
m=a[j][i]+k*10;
a[j][i]=m/(j+1);
k=m%(j+1);
}
while(!a[j][l])
l--;
a[j][0]=l;
}
}
int main()
{
C();
int n;
while(scanf("%d",&n)!=EOF)
{
int i;
for(i=a[n][0];i>=1;i--)
printf("%d",a[n][i]);
printf("\n");
}
return 0;
}
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原文地址:http://www.cnblogs.com/-lgh/p/5012631.html