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Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4981 Accepted Submission(s): 2085
#include<cstdio>
#include<cstring>
int a[390];
int n,m,leap;
void bfact(int n)
{
int i,j;
for(i=2; i<=n; i++)
{
if(leap&&i==m+1) continue;
int c=0,s;
for(j=0; j<380; j++)
{
s=i*a[j]+c;
a[j]=s%10;
c=s/10;
}
}
}
void bx(int n)
{
int j;
int s,c=0;
for(j=0; j<380; j++)
{
s=n*a[j]+c;
a[j]=s%10;
c=s/10;
}
}
int main()
{
//freopen("case.in","r",stdin);
int i,j,c=1;
while(scanf("%d%d",&m,&n)!=-1)
{
if(!m&&!n) break;
leap=0;
memset(a,0,sizeof(a));
a[0]=1;
printf("Test #%d:\n",c);
if(m<n)
{
printf("0\n");
c++;
continue;
}
if((m-n+1)%(m+1)==0)
bx((m-n+1)/(m+1));
else
{
bx(m-n+1);
leap=1;
}
bfact(n+m);
for(i=380; i>=0; i--)
if(a[i]) break;
for(j=i; j>=0; j--)
printf("%d",a[j]);
printf("\n");
c++;
}
}
//从网上找到了公式 即:结果等于 (m+n)!*(m-n+1)/(m+1)
//那么这题就是高精度问题了
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原文地址:http://www.cnblogs.com/orchidzjl/p/4314408.html