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  • # [SDOI2019]移动金币 阶梯博弈 dp

    [SDOI移动金币

    链接

    vijos

    思路

    阶梯博弈,dp统计.
    参见wxyww

    代码

    #include <bits/stdc++.h>
    using namespace std;
    const int N = 2e5 + 7, mod = 1e9 + 9;
    int read() {
        int x = 0, f = 1; char s = getchar();
        for (;s > '9' || s < '0'; s = getchar()) if (s == '-') f = -1;
        for (;s >= '0' && s <= '9'; s = getchar()) x = x * 10 + s - '0';
        return x * f;
    }
    int n, m, f[20][N], jc[N], inv[N];
    int q_pow(int a, int b) {
        int ans = 1;
        while (b) {
            if (b & 1) ans = 1LL * a * ans % mod;
            a = 1LL * a * a % mod;
            b >>= 1;
        }
        return ans;
    }
    int C(int n, int m) {return 1LL * jc[n] * inv[m] % mod * inv[n-m] % mod;}
    int main() {
        n = read(), m = read();
        jc[0] = inv[0] = jc[1] = inv[1] = 1;
        for (int i = 2; i <= n; ++i) {
            jc[i] = 1LL * jc[i-1] * i % mod;
            inv[i] = q_pow(jc[i], mod - 2);
        }
        int ans = C(n, m);
        n -= m;
        int num = (m + 1) >> 1;
        f[0][0] = 1;
        for (int i = 1; i <= 19; ++i) {
            for (int j = 0; j <= n; ++j) {
                for (int k = 0; (k << i - 1) <= j && k <= num; k += 2) {
                    f[i][j] += 1LL * f[i-1][j - (k << i - 1)] * C(num, k) % mod;
                    f[i][j] %= mod;
                }
            }
        }
        for (int i = 0; i <= n; ++i) {
            ans -= 1LL * f[19][i] * C(n + m / 2 - i, m / 2) % mod;
            ans = (ans + mod) % mod; 
        }
        printf("%lld", ans);
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/dsrdsr/p/10975570.html
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