组合数1000内

组合数 C<1000；
利用递推快速得到C[n][m]=C[n-1][m-1]+C[n-1][m];

int C[1001][1001];
void Init()
{  C[0][0]=1;
   fin(1,1000)
   {
    C[i][0]=1;
   }
   fin(1,1000)
   {
    for(int j=1;j<=i;j++)
    {   
     C[i][j]=(C[i-1][j]%md+C[i-1][j-1]%md)%md;
    }
   }
}

https://www.luogu.org/problemnew/show/P1313
附上一到练手题AC代码

#include<bits/stdc++.h>
using namespace std;
#define ll long long int
#define fin(a,n) for(int i=a;i<=n;i++)
const int md=10007;
const int maxn=1e3+10;
int C[1001][1001];
void Init()
{  C[0][0]=1;
   fin(1,1000)
   {
    C[i][0]=1;
   }
   fin(1,1000)
   {
    for(int j=1;j<=i;j++)
    {   
     C[i][j]=(C[i-1][j]%md+C[i-1][j-1]%md)%md;
    }
   }
}
ll ksm(ll a,ll b)
{ 
 ll ans=1;
 a=a%md;
 while(b)
 {
  if(b%2==1)ans=(ans*a)%md;
  b/=2;
  a=(a*a)%md;
 }
 return ans;
}
int main()
{   ll a,b,k,n,m;
    Init();
 scanf("%lld %lld %lld %lld %lld",&a,&b,&k,&n,&m);
 ll ans=(ksm(a,n)*ksm(b,m))%md;
    ll tans=(ans%md*(C[k][n]%md))%md;
    printf("%lld",tans);
}