HDU-6114

虽然这是一道水题，不过这道c(n,m)%mod 的模板值得记录

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mod=1e9+7;
const int maxn = 2000 + 5;

int F[maxn], Finv[maxn], inv[maxn];
void init()
{
    inv[1] = 1;
    for(int i = 2; i < maxn; i ++)
    {
        inv[i] = (mod - mod / i) * 1ll * inv[mod % i] % mod;
    }
    F[0] = Finv[0] = 1;
    for(int i = 1; i < maxn; i ++)
    {
        F[i] = F[i-1] * 1ll * i % mod;
        Finv[i] = Finv[i-1] * 1ll * inv[i] % mod;
    }
}
int comb(int n, int m) //comb(n, m)就是C(n, m)
{
    if(m < 0 || m > n) return 0;
    return F[n] * 1ll * Finv[n - m] % mod * Finv[m] % mod;
}
int main()
{
    int t;
    init();
    scanf("%d",&t);
    while(t--)
    {
        int n,m;
        scanf("%d%d",&n,&m);
        printf("%d\n",comb(max(n,m),min(n,m)));
    }
}

 

也可以用dp做，记录一个递推公式 c(n,m)=(c(n-1,m-1)+c(n-1,m))%mod;

#include<cstdio>
#include<cstring>
#include<algorithm>
#define LL long long
using namespace std;
const int mod=1000000007;
int T,n,m,ans;
LL c[1005][1005];
LL C(int x,int y)
{
    if(y==x||y==0)return 1;
    if(y==1)return x;
    if(c[x][y])return c[x][y]%mod;
    c[x][y]=(C(x-1,y-1)+C(x-1,y))%mod;
    return c[x][y];
}
int main()
{
    scanf("%d",&T);
    while(T--){
        scanf("%d %d",&n,&m);
        printf("%I64d\n",C(max(n,m),min(n,m)));
    }
    return 0;
}