南阳理工ACM1076--方案数量

<span style="font-size:18px;">#include <stdio.h>

int main()
{
long long a[32][32];
int i,j;
int m,n;
a[0][0]=0;
for (i=1;i<32;i++)
{
a[0][i]=1;
a[i][0]=1;
}
for (i=1;i<32;i++)
{
for (j=1;j<32;j++)
{
a[i][j] = a[i][j-1]+a[i-1][j];
}

}
while (scanf("%d%d",&m,&n)!=EOF&&(m!=0||n!=0))
printf("%lld\n",a[m]
);
return 0;
}
</span>