【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 
#define enter putchar('\n')
#define space putchar(' ')
#define pii pair
#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
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
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 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;
}

转载于:https://www.cnblogs.com/ivorysi/p/9572797.html