2019牛客多校 Round9

Solved:3

Rank:181

 

H Cutting Bamboos

这个东西好像叫整体二分

#include 
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;
}
Cutting Bamboos

 

转载于:https://www.cnblogs.com/lwqq3/p/11364117.html

你可能感兴趣的:(2019牛客多校 Round9)