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  • [AHOI 2009]chess 中国象棋

    Description

    题库链接

    给你一张 (N imes M) 的棋盘。要求每行每列最多放两个棋子,问总方案数。

    (1leq N,Mleq 100)

    Solution

    (f_{i,j,k}) 为前 (i) 行还剩 (j) 行可以放 (1) 个棋子, (k) 行放两个棋子的方案数。组合数学乱搞就好了。

    Code

    //It is made by Awson on 2018.3.17
    #include <bits/stdc++.h>
    #define LL long long
    #define dob complex<double>
    #define Abs(a) ((a) < 0 ? (-(a)) : (a))
    #define Max(a, b) ((a) > (b) ? (a) : (b))
    #define Min(a, b) ((a) < (b) ? (a) : (b))
    #define Swap(a, b) ((a) ^= (b), (b) ^= (a), (a) ^= (b))
    #define writeln(x) (write(x), putchar('
    '))
    #define lowbit(x) ((x)&(-(x)))
    using namespace std;
    const int N = 100, yzh = 9999973;
    void read(int &x) {
        char ch; bool flag = 0;
        for (ch = getchar(); !isdigit(ch) && ((flag |= (ch == '-')) || 1); ch = getchar());
        for (x = 0; isdigit(ch); x = (x<<1)+(x<<3)+ch-48, ch = getchar());
        x *= 1-2*flag;
    }
    void print(unsigned LL x) {if (x > 9) print(x/10); putchar(x%10+48); }
    void write(unsigned LL x) {if (x < 0) putchar('-'); print(Abs(x)); }
    
    int f[N+5][N+5][N+5], n, m, C[N+5][N+5];
    
    void work() {
        for (int i = 0; i <= N; i++) {
        C[i][0] = 1;
        for (int j = 1; j <= i; j++) C[i][j] = (C[i-1][j]+C[i-1][j-1])%yzh;
        }
        read(n), read(m);
        f[0][0][m] = 1;
        for (int i = 1; i <= n; i++)
        for (int j = 0; j <= m; j++)
            for (int k = 0; k <= m; k++)
            for (int p = 0; p <= 2; p++)
                for (int q = 0; q+p <= 2; q++)
                    (f[i][j+q][k] += 1ll*f[i-1][j+p][k+q]*C[j+p][p]%yzh*C[k+q][q]%yzh) %= yzh;
        int ans = 0;
        for (int i = 0; i <= m; i++) for (int j = 0; j <= m; j++) (ans += f[n][i][j]) %= yzh;
        writeln(ans);
    }
    int main() {
        work(); return 0;
    }
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  • 原文地址:https://www.cnblogs.com/NaVi-Awson/p/8590824.html
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