2019ICPC 南昌邀请赛 B - Polynomial

\(2019ICPC\) 南昌邀请赛 \(B - Polynomial\)

Define \(f_{(x)} = a_0+a_1 \times x^1 + a_2 \times x^2 +…+ a_n \times x^n\) the number maybe tremendous huge, so take the module of \(9999991\).

For a polynomial, \(XH\) doesn’t know any \(a_i\) , but he knows the value of \(f_{(0)} , f_{(1)}, f_{(2)}…f_{(n)}\). He wants to calculate \(\displaystyle\sum_{i=L}^R f_{(i)}= mod\ 9999991\).

Input

The first line contains a number \(T ( 1 \le T \le 5 )\) ——the number of test cases.

For each test case. The first line contains two integers \(n (1 \le n \le 1000)\) and \(m(1 \le m \le 2000)\) .

The second line contains \(n+1\) integers,representing the value of \(f_{(0)} , f_{(1)}, f_{(2)}…f_{(n)}\)

The next mm lines, each line contains two integers \(L, R\).

\(( 1 \le L \le R \le 9999990)\)

Output

For each test case, output mm lines, each line contains one integer, the value of \(\displaystyle\sum_{i=L}^R f_{(i)}= mod\ 9999991\).

输出时每行末尾的多余空格，不影响答案正确性

样例输入

1 
3 2
1 10 49 142
6 7 
95000 100000 

样例输出

2519 
1895570 

题解

已知 \(f_x\) ，假设一个 \(S_x=\sum_{i=0}^{x}f_x\) ，那么 \(\displaystyle\sum_{i=L}^R f_{(i)}=S_R-S_{L-1}\)

对于构造 \(S_x\)

\(1^n+2^n+...+x^n\) 是一个 \(x^{n+1}\) 级的式子，具体资料查阅 https://pan.baidu.com/s/1SqZMrZLuOqIp1HQVQejK7g

那么需要在 $ S_x$ 需要在 \(f_x\) 基础上新添加一个项 \(f_{n+1}\)，使用拉格朗日插值法进行求解

之后使用递推式 \(S_i=S_{i-1}+f_i\) 求解出一个新的 \(n+1\) 次函数的 \(n+2\) 个点

详细的拉格朗日插值法讲解 https://www.cnblogs.com/crashed/p/13124448.html

注意

• 如果正常的求解会 \(TLE\) ，一些操作需要进行提前处理，减少时间开支。

代码

#include <cstdio>
#include <iostream>
using namespace std;
typedef long long ll;
const ll mod = 9999991;
const int maxn = 1e3+55;
const int maxxn = 1e7+55;
ll a[maxn];
ll fac[maxn];
ll sub[maxn];
ll pow(ll a,ll b){
	ll ret=1;
	while(b){
		if(b&1)ret=ret*a%mod;
		b>>=1;
		a=a*a%mod;
	}
	return ret;
}
ll slove(ll x,int n){
	ll ans=0;
	if(x<=n)return a[x];
	ll xx=1;
	for(int i=0;i<=n;++i){
		xx=xx*(x-i)%mod;
	}
	for(int i=0;i<=n;++i){
		ll yy=fac[i]*fac[n-i]%mod;
		if((n-i)%2)yy=-yy;
		ans+=(((a[i]*xx)%mod)*sub[i]%mod)*pow(x-i,mod-2)%mod;
		ans=(ans%mod+mod)%mod;
	}
	return ans;
}
void init(int n){
	fac[0]=1;
	for(int i=1;i<=n;++i){
		fac[i]=fac[i-1]*i%mod;
	}
	for(int i=0;i<=n;++i){
		ll yy=fac[i]*fac[n-i]%mod;
		if((n-i)%2)yy=-yy;
		sub[i]=pow(yy,mod-2);
	}
}
int main(){
	int t;
	scanf("%d",&t);
	while(t--){
		int n,m;
		scanf("%d%d",&n,&m);
		init(n);
		for(int i=0;i<=n;++i){
			scanf("%lld",&a[i]);
		}
		a[n+1]=slove((ll)(n+1),n);
		init(n+1);
		for(int i=1;i<=n+1;++i){
			a[i]+=a[i-1];
			a[i]%=mod;
		}
		while(m--){
			ll l,r;
			scanf("%lld%lld",&l,&r);
			printf("%lld\n",((slove(r,n+1)-slove(l-1,n+1))%mod+mod)%mod);
		}
	}
	return 0;
}