1 条题解

  • 0
    @ 2022-8-17 22:32:48
    #include <bits/stdc++.h>
    using namespace std;
    typedef long long LL;
    const int INF = 0x3f3f3f3f;
    const LL mod = 1e9 + 7;
    const int N = 500005;
    
    LL pow_mod(LL x, LL n) {
        LL res = 1;
        while (n) {
            if (n & 1) res = res * x % mod;
            n >>= 1;
            x = x * x % mod;
        }
        return res;
    }
    
    LL fac[N];
    // 计算 1^m+2^m+3^m+...+n^m
    LL cal(LL n, LL m) {
        LL res = 0;
        if (n <= m + 2) {
            for (int i = 1; i <= n; i++) {
                res = (res + pow_mod(i, m)) % mod;
            }
        } else {
            fac[0] = 1;
            for (int i = 1; i <= m + 1; i++) {
                fac[i] = fac[i - 1] * i % mod;
            }
            LL t = 1;
            for (int i = 1; i <= m + 2; i++) {
                t = t * (n - i) % mod;
            }
            LL y = 0;
            int flag = (m + 2) % 2 ? 1 : -1;
            for (int i = 1; i <= m + 2; i++) {
                y = (y + pow_mod(i, m)) % mod;
                res += flag * y * t % mod * pow_mod(n - i, mod - 2) % mod * pow_mod(fac[i - 1] * fac[m + 2 - i] % mod, mod - 2) % mod;
                flag = -flag;
            }
            res = (res % mod + mod) % mod;
        }
        return res;
    }
    
    int w[20], e[20], l[N][20], r[N][20];
    int a[20], b[20], h[20];
    LL f[20][20];
    int c[N], d[N];
    int n, m;
    
    int main() {
        scanf("%d%d", &n, &m);
        for (int i = 1; i <= m; i++) {
            scanf("%d", &w[i]);
        }
        for (int i = 1; i <= n; i++) {
            scanf("%d%d", &c[i], &d[i]);
            e[c[i]] += d[i];
            for (int j = 1; j <= m; j++) {
                l[i][j] = l[i - 1][j];
                r[i][j] = r[i - 1][j];
            }
            l[i][c[i]] = min(l[i][c[i]], e[c[i]]);
            r[i][c[i]] = max(r[i][c[i]], e[c[i]]);
        }
        bool lose = 1;
        for (int i = 1; i <= m; i++) {
            if (e[i] != 0 || r[n][i] - l[n][i] >= w[i]) {
                lose = 0;
            }
        }
        if (lose) return puts("-1"), 0;
        for (int j = 1; j <= m; j++) {
            a[j] = w[j] - (r[n][j] - l[n][j]);
        }
        LL ans = 0;
        // 第一轮贡献
        for (int i = 0; i <= n; i++) {
            LL s = 1;
            for (int j = 1; j <= m; j++) {
                s = s * max(0, (w[j] - (r[i][j] - l[i][j]))) % mod;
            }
            ans = (ans + s) % mod;
        }
        // 第二轮的死亡范围更新
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= m; j++) {
                r[i][j] = max(0, r[i][j] + e[j] - r[n][j]);
                l[i][j] = min(0, l[i][j] + e[j] - l[n][j]);
            }
        }
        for (int j = 1; j <= m; j++) {
            b[j] = r[n][j] - l[n][j];
        }
        // 第二轮开始的贡献
    
        int last = -1;
        for (int i = 1; i <= n; i++) {
            for (int j = 0; j <= m; j++) f[0][j] = 0;
            f[0][0] = 1;
            int t = INF;
            for (int j = 1; j <= m; j++) {
                int x = a[j] - r[i][j] + l[i][j];
                if (x <= 0) goto M1;  // 第二轮就暴毙了
                if (b[j] > 0) t = min(t, x / b[j]);
                for (int k = 0; k <= m; k++) {
                    f[j][k] = f[j - 1][k] * x % mod;
                    if (k > 0)
                        f[j][k] = (f[j][k] + f[j - 1][k - 1] * -b[j]) % mod;
                }
            }
            ans += f[m][0] * (t + 1) % mod;
            if (t != last) {
                last = t;
                for (int j = 1; j <= m; j++) h[j] = cal(t, j);
            }
            for (int j = 1; j <= m; j++) {
                ans += h[j] * f[m][j] % mod;
            }
        }
    M1:;
        ans = (ans % mod + mod) % mod;
        printf("%lld\n", ans);
        return 0;///test
    }
    
    • 1

    信息

    ID
    89
    时间
    1000ms
    内存
    512MiB
    难度
    9
    标签
    递交数
    9
    已通过
    5
    上传者