2019牛客暑期多校训练营（第九场）

题号	标题	已通过代码	题解/讨论	通过率	团队的状态
A	The power of Fibonacci	点击查看	进入讨论	34/152	未通过
B	Quadratic equation	点击查看	进入讨论	329/741	通过
C	Inversions of all permutations	点击查看	进入讨论	21/34	未通过
D	Knapsack Cryptosystem	点击查看	进入讨论	533/2123	通过
E	All men are brothers	点击查看	进入讨论	366/938	通过
F	Birthday Reminders	点击查看	进入讨论	5/11	未通过
G	Checkers	点击查看	进入讨论	5/8	未通过
H	Cutting Bamboos	点击查看	进入讨论	145/628	已补
I	KM and M	点击查看	进入讨论	15/253	5yisdf
J	Symmetrical Painting	点击查看	进入讨论	195/797	已补

B

https://www.cnblogs.com/1625--H/p/11382466.html

D

折半搜索，36/2 = 18，状压表示\(2^{18}\) 个结果，然后前一半与后一半去匹配。

#include <bits/stdc++.h>
using namespace std;
int n;
long long s;
long long a[70];
vector<long long> L, R;
vector<pair<long long, long long> > Ls, Rs;
void gao(const vector<long long>& A, vector<pair<long long, long long> >& As) {
    int n = A.size();
    for (int i = 0; i < 1 << n; i++) {
        long long s = 0;
        for (int j = 0; j < n; j++) {
            if (i >> j & 1) {
                s += A[j];
            }
        }
        As.push_back(make_pair(s, i));
    }
    sort(As.begin(), As.end());
}
long long work() {
    int i = 0, j = Rs.size() - 1;
    while (i < Ls.size() && j >= 0) {
        if (Ls[i].first + Rs[j].first < s) {
            i++;
        } else if (Ls[i].first + Rs[j].first > s) {
            j--;
        } else {
            return Rs[j].second << L.size() | Ls[i].second;
        }
    }
    assert(false);
}
int main() {
    cin >> n >> s;
    for (int i = 0; i < n; i++) {
        cin >> a[i];
        if (i < n / 2) {
            L.push_back(a[i]);
        } else {
            R.push_back(a[i]);
        }
    }
    gao(L, Ls);
    gao(R, Rs);
    long long ans = work();
    for (int i = 0; i < n; i++) {
        printf("%d", (int)(ans >> i & 1));
    }
    printf("\n");
    return 0;
}

E

https://www.cnblogs.com/1625--H/p/11359772.html

H

队友补的

#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2e5 + 5;
int n, m, cnt, len;
int a[MAXN];
int b[MAXN];
int t[MAXN];
int num[MAXN << 5];
int ls[MAXN << 5];
int rs[MAXN << 5];
double sum[MAXN << 5];
int build(int l, int r) {
    int rt = ++cnt;
    sum[0] = 0, num[0] = 0;
    int mid = l + r >> 1;
    if(l < r) {
        ls[rt] = build(l, mid);
        rs[rt] = build(mid + 1, r);
    }
    return rt;
}
 
int add(int o, int l, int r, int k, int v) {
    int rt = ++cnt;
    ls[rt] = ls[o]; rs[rt] = rs[o]; sum[rt] = sum[o] + 1.0 * v; num[rt] = num[o] + 1;
 
    int mid = l + r >> 1;
    if(l < r)
    if(k <= mid) ls[rt] = add(ls[o], l, mid, k, v);
    else rs[rt] = add(rs[o], mid + 1, r, k, v);
    return rt;
}
 
double query(int ql, int qr, int l, int r, double k, int tot) {
    if(l == r) return  k / (1.0 * (num[qr] - num[ql] + tot));
 
    int mid = l + r >> 1;
    double lsum = sum[ls[qr]] - sum[ls[ql]];
    double rsum = 1.0 * mid * (num[rs[qr]] - num[rs[ql]] + tot);
    if(lsum + rsum > k) return query(ls[ql], ls[qr], l, mid, k, tot + num[rs[qr]] - num[rs[ql]]);
    else return query(rs[ql], rs[qr], mid + 1, r, k - lsum, tot);
}
 
int main() {
    cnt = 0;
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; i++) scanf("%d", &a[i]), b[i] = a[i];
    sort(b + 1, b + 1 + n);
    len = unique(b + 1, b + 1 + n) - b - 1;
 
    t[0] = build(0, 100000);
    for(int i = 1; i <= n; i++) t[i] = add(t[i - 1], 0, 100000, a[i], a[i]);
 
    while(m--) {
        int l, r, x, y;
        scanf("%d%d%d%d", &l, &r, &x, &y);
        double tmp = 1.0 * (sum[t[r]] - sum[t[l - 1]]) / (1.0 * y);
        printf("%.8lf\n", query(t[l - 1], t[r], 0, 100000, tmp * (y - x), 0));
    }
    return 0;
}

ps:主席树也不是很难，得赶快学了。

J

https://www.cnblogs.com/1625--H/p/11383636.html