FZU 8月有奖月赛A Daxia Wzc's problem (Lucas)

Problem A Daxia & Wzc's problem

Accept: 42    Submit: 228
Time Limit: 1000 mSec    Memory Limit : 32768 KB

 Problem Description

Daxia在2016年5月期间去瑞士度蜜月,顺便拜访了Wzc,Wzc给他出了一个问题：

 

Wzc给Daxia等差数列A(0),告诉Daxia首项a和公差d;

 

首先让Daxia求出数列A(0)前n项和,得到新数列A(1);

 

然后让Daxia求出数列A(1)前n项和,得到新数列A(2);

 

接着让Daxia求出数列A(2)前n项和,得到新数列A(3);

 

...

 

最后让Daxia求出数列A(m-1)前n项和,得到新数列A(m);

 Input

测试包含多组数据,每组一行,包含四个正整数a(0<=a<=100),d(0<d<=100),m(0<m<=1000),i(1<=i<=1000000000).

 Output

每组数据输出一行整数,数列A(m)的第i项mod1000000007的值.

 Sample Input

1 1 3 4

 Sample Output

35

 Hint

A(0): 1 2 3 4

 

A(1): 1 3 6 10

 

A(2): 1 4 10 20

 

A(3): 1 5 15 35

 

So the 4th of A(3) is 35.
Cached at 2016-08-17 19:08:15.

 

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草稿纸上手写一下

a1  a1+d  a1+2d  a1+3d...

a1  2a1+d  3a1+3d  4a1+6d...

a1  3a1+d  6a1+4d  10a1+10d...

...

可以发现这个是一个类似杨辉三角的东西，大概就是C(n, m)这样算的。

然后就用Lucas就行了

//#pragma comment(linker, "/STACK:102400000, 102400000")
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cmath>
#include <ctime>
#include <list>
#include <set>
#include <map>
using namespace std;
typedef long long LL;
typedef pair <int, int> P;
const int N = 1e3 + 5;
LL mod = 1e9 + 7;
LL f[N];

LL Pow(LL a , LL n , LL mod) {
    LL res = 1;
    while(n) {
        if(n & 1)
            res = res * a % mod;
        a = a * a % mod;
        n >>= 1;
    }
    return res;
}

LL Comb(LL a , LL b , LL mod) {
    if(a < b) {
        return 0;
    }
    if(a == b) {
        return 1;
    }
    LL ca = 1;
    for(LL i = 0 ; i < b ; i++) {
        ca = (a - i) % mod * ca % mod;
    }
    return (ca * f[b]) % mod;
}

LL Lucas(LL n , LL m , LL mod) {
    LL ans = 1;
    while(m && n && ans) {
        ans = (ans * Comb(n % mod , m % mod , mod)) % mod;
        n /= mod;
        m /= mod;
    }
    return ans;
}

int main()
{
    f[0] = 1;
    for(LL j = 1; j < N; ++j) {
        f[j] = j * f[j - 1] % mod; //阶乘
    }
    for(LL j = 0; j < N; ++j) {
        f[j] = Pow(f[j], mod - 2, mod); //逆元
    }
    LL a, b, m, i;
    while(cin >> a >> b >> m >> i) {
        if(i == 1) {
            cout << a << endl;
            continue;
        }
        LL x = Lucas(m + i - 1, m, mod) * a % mod;
        LL y = Lucas(m + 1 + i - 2, m + 1, mod) * b % mod;
        cout << (x + y) % mod << endl;
    }
    return 0;
}