FOJ有奖月赛-2016年8月（daxia专场之过四题方有奖）

http://acm.fzu.edu.cn/contest/list.php?cid=152

主要是a题, lucas定理, 就这一版能过..  记录一下代码, 另外两个最短路  一个模拟,没什么记录价值.

//#define txtout
//#define debug
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
#define mt(a,b) memset(a,b,sizeof(a))
using namespace std;
typedef long long LL;
const double pi=acos(-1.0);
const double eps=1e-8;
const int inf=0x3f3f3f3f;
//const int M=1e7+10;

LL pow_mod (LL a, LL n, LL p) {
    LL ans = 1,t = a;
    while (n) {
        if (n & 1) {
            ans = ans * t % p;
        }
        t = t * t % p;
        n >>= 1;
    }
    return ans;
}
LL cal (LL n, LL m, LL p) {
    if(m > n-m) m = n - m;
    LL ans = 1;
    for (int i = 1; i <= m; i++) {
        ans = ans * (n - i + 1) % p;
        int a = pow_mod(i,p-2,p);
        ans = ans * a % p;
    }
    return ans;
}
LL com_mod (LL n,LL m,LL p) {
    if (n < m)return 0;
    return cal(n,m,p);
}
LL lucas (LL n, LL m, LL p) {
    LL r = 1;
    while(n && m && r) {
        r = r *com_mod(n%p, m%p, p) % p;
        n /= p;
        m /= p;
    }
    return r;
}

LL a,d,m,n;
//int c[M];
//LL C(int n,int m) {
//    LL result=1;
//    for(int i=0; i<m; i++) {
//        result*=(n-i);
//    }
//    for(int i=0; i<m; i++) {
//        result/=(1+i);
//    }
//    return result;
//}
const LL MOD=1000000007;
LL solve() {
    if(n==1){
        return a;
    }
    LL answer=lucas(m+n-1,n-1,MOD)*a%MOD;
    if(n-2>=0) answer+=lucas(m+n-1,n-2,MOD)*d%MOD;
    answer%=MOD;
    return answer;
}
int main() {
#ifdef txtout
    freopen("in.txt","r",stdin);
    freopen("out.txt","w",stdout);
#endif // txtout
//    num=sieve(MAXN);
    while(~scanf("%I64d%I64d%I64d%I64d",&a,&d,&m,&n)) {
        printf("%I64d
",solve());
    }
    return 0;
}

end