无限手套

http://www.nowcoder.com/questionTerminal/e15f205e3c3a46cba74bfabc343979ba

牛客无限手套

输入描述:

第一行一个正整数m表示宝石的种类（1<=m<=1000）接下来M行，每行两个正整数ai, bi（0<=ai, bi<=10^9）接下来一行正整数q，共有q次询问（1<=q<=1000）接下来q行每行一个正整数n询问如果无限手套可以安装n个宝石则力量之和是多少。（1<=n<=10000）

输出描述:

一共q行,每行一个正整数表示答案。答案对998244353取模。

示例1

输入

输出

74
193

题解

考虑对每一个宝石的生成函数为

$\sum_{k=0}^{\infty}(a_ik^2+b_ik+1)x^k$

化简过程：

$\begin{aligned} &\sum_{k=0}^{\infty}(a_ik^2+b_ik+1)x^k\\ =&\sum_{k=0}^{\infty}x^k+b_i\sum_{k=0}^{\infty}kx^k+a_i\sum_{k=0}^{\infty}k^2x^k\\ =&\cfrac{1}{1-x}+\cfrac{b_ix}{(1-x)^2}+\cfrac{a_ix(1+x)}{(1-x)^3}\\ =&\cfrac{x^2(a_i-b_i+1)+x(a_i+b_i-2)+1}{(1-x)^3} \end{aligned}$

其中也许令你迷惑的是

$\displaystyle\sum_{k=0}^{\infty}k^2x^k=\cfrac{x(1+x)}{(1-x)^3}$

推导过程如下：

$\begin{aligned} \sum_{k=0}^{\infty}x^k&=\cfrac{1}{1-x}\ \ (|x|<1)\\ \implies\ \cfrac{d}{dx}(\sum_{k=0}^{\infty}x^k)&=\cfrac{1}{(1-x)^2}\\ \sum_{k=0}^{\infty}kx^{k-1}&=\cfrac{1}{(1-x)^2}\\ \sum_{k=0}^{\infty}kx^{k}&=\cfrac{x}{(1-x)^2}\\ \implies\ \cfrac{d}{dx}(\sum_{k=0}^{\infty}kx^k)&=\cfrac{(1-x)^2+2x(1-x)}{(1-x)^4}\\ \sum_{k=0}^{\infty}k^2x^{k-1}&=\cfrac{1+x}{(1-x)^3}\\ \implies\ \sum_{k=0}^{\infty}k^2x^k&=\cfrac{x(1+x)}{(1-x)^3} \end{aligned}$

m种宝石的生成函数相乘

$\cfrac{\displaystyle\prod_{i=1}^{m}[(a_i-b_i+1)x^2+(a_i+b_i-2)x+1]}{(1-x)^{3m}}$

对于 $\cfrac{1}{(1-x)^{3m}}$ 我们用幂级数展开，展开方法为：

$\begin{aligned} (1-x)^{-n}&=(-x)^0+(-n)(-x)^1+\cfrac{(-n)(-n-1)}{2!}(-x)^2+\cdots\\ &=1+nx+\cfrac{n(n+1)}{2!}x^2+\cfrac{n(n+1)(n+2)}{3!}x^3+\cdots\\ &=1+nx+C_{n+1}^{2}x^2+C_{n+2}^{3}x^3+\cdots \end{aligned}$

于是式子转化为两个多项式相乘
最后 $x^n$ 的系数即为所求

代码

// #include <bits/stdc++.h>
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <queue>
#include <cmath>
#include <string>
#include <cstring>
#include <map>
#include <unordered_map>
#include <set>
#include <vector>
#include <assert.h>
#include <cmath>
#include <ctime>
using namespace std;
#define me(x,y) memset((x),(y),sizeof (x))
#define MIN(x,y) ((x) < (y) ? (x) : (y))
#define MAX(x,y) ((x) > (y) ? (x) : (y))
#define SGN(x) ((x)>0?1:((x)<0?-1:0))
#define ABS(x) ((x)>0?(x):-(x))
// #define int __int128_t

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

const int maxn = 1e5+10;
const int inf = __INT32_MAX__;
const ll INF = __LONG_LONG_MAX__;
const ll MOD = 998244353;
const double eps = 1e-8;
const double pi = std::acos(-1);
const string cars[] = {"🚗","🚕","🚙"};

ll f[maxn],finv[maxn],inv[maxn];
void init(){
    inv[1] = 1;
    for(int i = 2; i < maxn;++i) inv[i] = 1ll*(MOD-MOD/i)*inv[MOD%i]%MOD;
    f[0] = finv[0] = 1;
    for(int i = 1; i < maxn; ++i){
        f[i] = f[i-1]*i%MOD;
        finv[i] = finv[i-1]*inv[i]%MOD;
    }
}
ll comb(int n,int m){
    if(m < 0 || m > n) return 0;
    return f[n]*finv[n-m]%MOD*finv[m]%MOD;
}

ll x[maxn];
int main(){
    ios::sync_with_stdio(false);
#ifndef ONLINE_JUDGE
    freopen("1in.in","r",stdin);
    freopen("1out.out","w",stdout);
#endif    
    init();
    int a,b,A,B,C,m;cin>>m;
    x[0] = 1;
    for(int i = 1; i <= m; ++i){
        cin>>a>>b;
        A = a-b+1,B = a+b-2,C = 1;
        A = (A%MOD+MOD)%MOD,B = (B%MOD+MOD)%MOD;
        for(int j = 2*i; j >= 2; --j) x[j] = (x[j]*C%MOD+x[j-1]*B%MOD+x[j-2]*A%MOD)%MOD;
        x[1] = (x[1]*C%MOD+x[0]*B%MOD)%MOD;
        x[0] = x[0]*C%MOD;
    }
    int q;cin>>q;
    while(q--){
        int n;cin>>n;
        ll sum = 0;
        for(int i = 0; i <= n; ++i) sum = (sum+comb(3*m+i-1,i)*x[n-i]%MOD)%MOD;
        cout<<sum<<endl;
    }
    return 0;
}