FZU Problem 2238 Daxia & Wzc's problem

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;

#include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
LL a,d,m,i;
const int MOD = 1000000007;
LL quick_pow (LL a,LL b,LL MOD)
{
    //求解 a^b%MOD的值
    LL base=a%MOD;
    LL ans=1; //相乘，所以这里是1
    while (b)
    {
        if (b&1)
        {
            ans=(ans*base)%MOD; //如果这里是很大的数据，就要用quick_mul
        }
        base=(base*base)%MOD;    //notice
        //注意这里,每次的base是自己base倍
        b>>=1;
    }
    return ans;
}
LL C (LL n,LL m,LL MOD)
{
    if (n<m) return 0; //防止sb地在循环，在lucas的时候
    if (n==m) return 1;
    LL ans1 = 1;
    LL ans2 = 1;
    LL mx=max(n-m,m); //这个也是必要的。能约就约最大的那个
    LL mi=n-mx;
    for (int i = 1; i <= mi; ++i)
    {
        ans1 = ans1*(mx+i)%MOD;
        ans2 = ans2*i%MOD;
    }
    return (ans1*quick_pow(ans2,MOD-2,MOD)%MOD); //这里放到最后进行,不然会很慢
}


void work ()
{
    if (i==1)
    {
        printf ("%lld\n",a);
        return;
    }
    LL ans = (C(m+i-1,m,MOD)*a%MOD+C(m+i-1,i-2,MOD)*d%MOD)%MOD;
    printf ("%lld\n",ans);
    return ;
}
int main()
{
#ifdef local
    freopen("data.txt","r",stdin);
#endif
    while (scanf("%lld%lld%lld%lld",&a,&d,&m,&i)!=EOF) work();
    return 0;
}