【LOJ】#2098. 「CQOI2015」多项式

题解

令x = x - t代换一下会发现

\(\sum_{i = 0}^{n}a_i (x + t)^i = \sum_{i = 0}^{n} b_{i} x^{i}\)

剩下的就需要写高精度爆算了……

代码

#include <bits/stdc++.h>
#define enter putchar('\n')
#define space putchar(' ')
#define pii pair<int,int>
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define eps 1e-8
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-') f = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        res = res * 10 - '0' + c;
        c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) out(x / 10);
    putchar('0' + x % 10);
}
int BASE = 100000000;
int len = 8;
struct Bignum {
	vector<int> v;
	Bignum(int64 x = 0) {
		*this = x;
	}
	Bignum operator = (int64 x) {
		v.clear();
		do {
			v.pb(x % BASE);
			x /= BASE;
		}while(x);
		return *this;
	}
	Bignum operator = (string str) {
		v.clear();int x;
		for(int i = str.length() ; i > 0 ; i -= len) {
			int st = max(0,i - len),ed = i;
			sscanf(str.substr(st,ed - st).c_str(),"%d",&x);
			v.pb(x);
		}
		return *this;
	}
	friend Bignum operator + (const Bignum &a,const Bignum &b) {
		Bignum c;c.v.clear();
		int x,g = 0,p = 0;
		while(1) {
			x = g;
			if(p < a.v.size()) x += a.v[p];
			if(p < b.v.size()) x += b.v[p];
			if(!x && p >= a.v.size() && p >= b.v.size()) break;
			g = x / BASE;
			c.v.pb(x % BASE);
			++p;
		}
		return c;
	}
	friend Bignum operator - (const Bignum &a,const Bignum &b) {
		Bignum c;c.v.clear();
		int x,g = 0,p = 0;
		while(1) {
			x = -g;g = 0;
			if(p < a.v.size()) x += a.v[p];
			if(p < b.v.size()) x -= b.v[p];
			if(!x && p >= a.v.size() && p >= b.v.size()) break;
			if(x < 0) {x += BASE;g = 1;}
			c.v.pb(x);
			++p;
		}
		return c;
	}
	friend Bignum operator * (const Bignum &a,const Bignum &b) {
		Bignum c;c.v.clear();
		c.v.resize(a.v.size() + b.v.size());
		int64 x,g = 0;
		for(int i = 0 ; i < a.v.size() ; ++i) {
			g = 0;
			for(int j = 0 ; j < b.v.size() ; ++j) {
				x = 1LL * a.v[i] * b.v[j] + g + c.v[i + j];
				c.v[i + j] = x % BASE;
				g = x / BASE;
			}
			int t = i + b.v.size();
			while(g) {
				x = g + c.v[t];
				c.v[t] = x % BASE;
				g = x / BASE;
				++t;
			}
		}
		for(int i = c.v.size() - 1 ; i > 0 ; --i) {
			if(!c.v[i]) c.v.pop_back();
			else break;
		}
		return c;
	}
	friend Bignum operator / (const Bignum &a,const int &d) {
		Bignum c;
		c.v.resize(a.v.size());
		int64 x = 0,t;
		for(int i = a.v.size() - 1 ; i >= 0 ; --i) {
			t = 1LL * x * BASE + a.v[i];
			c.v[i] = t / d;
			x = t % d;
		}
		for(int i = c.v.size() - 1 ; i > 0 ; --i) {
			if(!c.v[i]) c.v.pop_back();
			else break;
		}
		return c;
	}
	void print() {
		int s = v.size() - 1;
		printf("%d",v[s]);
		--s;
		for(int i = s ; i >= 0 ; --i) {
			printf("%08d",v[i]);
		}
	}
}N,M,T,C,B,tmp;
string s[4];
struct Matrix {
	int f[2][2];
	Matrix(){memset(f,0,sizeof(f));}
	friend Matrix operator * (const Matrix &a,const Matrix &b) {
		Matrix c;
		for(int i = 0 ; i <= 1 ; ++i) {
			for(int j = 0 ; j <= 1 ; ++j) {
				for(int k = 0 ; k <= 1 ; ++k) {
					c.f[i][j] = (c.f[i][j] + a.f[i][k] * b.f[k][j]) % 3389;
				}
			}
		}
		return c;
	}
}A[15],ans;
int64 a[15],num[100005];
int64 fpow(int64 x,int64 c) {
	int64 res = 1,t = x;
	while(c) {
		if(c & 1) res = 1LL * res * t % 3389;
		t = 1LL * t * t % 3389;
		c >>= 1;
	}
	return res;
}
int main() {
#ifdef ivorysi
	freopen("f1.in","r",stdin);
#endif
	cin>>s[0]>>s[1]>>s[2];
	N = s[0];T = s[1],M = s[2];
	A[0].f[0][0] = A[0].f[1][1] = 1;
	A[1].f[0][0] = 1234,A[1].f[0][1] = 5678 % 3389,A[1].f[1][1] = 1;
	ans = A[0];
	for(int i = 2 ; i <= 10 ; ++i) A[i] = A[i - 1] * A[1];
	int c = (N - M).v[0];
	for(int i = 0 ; i < s[0].length() ; ++i) {
		Matrix t = A[0];
		for(int j = 1 ; j <= 10 ; ++j) t = t * ans;
		ans = t * A[s[0][i] - '0'];
	}
	a[0] = (ans.f[0][0] + ans.f[0][1]) % 3389;
	int64 inv = fpow(1234,3389 - 2);
	for(int i = 1 ; i <= c ; ++i) a[i] = 1LL * (a[i - 1] - 5678 + 3389 * 2) * inv % 3389;
	tmp = C = 1;
	for(int i = c; i >= 0 ; --i) {
		B = B + tmp * C * a[i];
		tmp = tmp * T;
		M = M + 1;
		C = C * M;
		C = C / (c - i + 1);
	}
	B.print();enter;
}