【LOJ】#2061. 「HAOI2016」放棋子

题解

水题，可惜要写高精度有点烦

一看障碍物的摆放方式和最后的答案没有关系，于是干脆不读了，直接二项式反演可以得到

设\(g_k\)为一种摆放方式恰好占了k个障碍物

\(f_k = \sum_{i = k}^{n} \binom{i}{k} g_{i}\)

可以得到

\(g_0 = \sum_{k = 0}^{n} (-1)^{k} \binom{k}{0} f_{i}\)

\(g_0 = \sum_{k = 0}^{n} (-1)^{k} \binom{n}{k} (n - k)!\)

拆开可以发现后面就是 \(\frac{n!}{k!}\)一个阶乘的后缀乘积

所以高精度只要实现高精乘低精，高精加，高精减就可以了

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 MAXN 200005
#define pb push_back
#define mp make_pair
//#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 + c - '0';
        c = getchar();
    }
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) out(x / 10);
    putchar('0' + x % 10);
}
const int BASE = 100000000;
const 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 > 0);
        return *this;
    }
    friend Bignum operator + (const Bignum &a,const Bignum &b) {
        Bignum c;c.v.clear();
        int p = 0,g = 0,x;
        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;
            c.v.pb(x % BASE);
            g = x / BASE;
            ++p;
        }
        return c;
    }
    friend Bignum operator - (const Bignum &a,const Bignum &b) {
        Bignum c;c.v.clear();
        int p = 0,g = 0,x;
        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) {g = 1;x += BASE;}
            c.v.pb(x);
            ++p;
        }
        return c;
    }
    friend Bignum operator * (const Bignum &a,int64 b) {
        Bignum c;c.v.clear();
        int s = a.v.size();
        int64 g = 0;
        for(int i = 0 ; i < s ; ++i) {
            int64 x = 1LL * a.v[i] * b + g;
            c.v.pb(x % BASE);
            g = x / BASE;
        }
        while(g) {
            c.v.pb(g % BASE);
            g /= BASE;
        }
        return c;
    }
    void print() {
        int s = v.size() - 1;
        printf("%d",v[s]);
        for(int i = s - 1 ; i >= 0 ; --i) {
            printf("%08d",v[i]);
        }
    }
}fac[205],ans;
int N;
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    read(N);
    if(N == 1) {puts("0");return 0;}
    fac[N + 1] = 1;
    for(int i = N ; i >= 1 ; --i) {
        fac[i] = fac[i + 1] * i;
    }

    for(int i = 0 ; i <= N ; i += 2) {
        ans = ans + fac[i + 1];
    }
    for(int i = 1 ; i <= N ; i += 2) {
        ans = ans - fac[i + 1];
    }
    ans.print();
    putchar('\n');
    return 0;
}