思路:建立一棵劃分樹,然後在區間裏面二分,尋找最大的小於a的數是第幾小的,以及尋找最大的小於等於b的數是第幾小的
兩個相減就是答案了。
#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define MAX 100050
#define m (l+r)>>1
int tree[30][MAX];
int toleft[30][MAX];
int sorted[MAX];
int n, k;
void build(int l, int r, int dep)
{
if (l == r)return;
int mid = m;
int cnt = mid - l + 1;
for (int i = l; i <= r; i++)
if (tree[dep][i] < sorted[mid])
cnt--;
int lpos = l;
int rpos = mid + 1;
for (int i = l; i <= r; i++)
{
if (tree[dep][i] < sorted[mid])
tree[dep + 1][lpos++] = tree[dep][i];
else
if (tree[dep][i] == sorted[mid] && cnt)
{
tree[dep + 1][lpos++] = tree[dep][i];
cnt--;
}
else
tree[dep + 1][rpos++] = tree[dep][i];
toleft[dep][i] = toleft[dep][l - 1] + lpos - l;
}
build(l, mid, dep + 1);
build(mid + 1, r, dep + 1);
}
int query(int L, int R, int l, int r, int dep, int k)
{
if (l == r)return tree[dep][l];
int cnt = toleft[dep][R] - toleft[dep][L - 1];
int mid = m;
if (cnt >= k)
{
int newL = l + toleft[dep][L - 1] - toleft[dep][l - 1];
int newR = newL + cnt - 1;
return query(newL, newR, l, mid, dep + 1, k);
}
else
{
int newR = R + toleft[dep][r] - toleft[dep][R];
int newL = newR - (R - L - cnt);
return query(newL, newR, mid + 1, r, dep + 1, k - cnt);
}
}
int gao(int s, int t, int a,int b)
{
int mid = 0;
int r = t - s + 1;
int l = 1;
int ans = 0;
while (l <= r)
{
mid = m;
int temp = query(s, t, 1, n, 0, mid);
if (temp < a)
{
ans = mid;
l = mid + 1;
}
else
r = mid - 1;
}
int ans2 = 0;
r = t - s + 1;
l = 1;
while (l <= r)
{
mid = m;
int temp = query(s, t, 1, n, 0, mid);
if (temp <= b)
{
ans2 = mid;
l = mid + 1;
}
else
r = mid - 1;
}
return ans2-ans;
}
int main()
{
int t;
scanf("%d", &t);
int icase = 1;
while (t--)
{
scanf("%d%d", &n, &k);
for (int i = 1; i <= n; i++)
{
scanf("%d", &tree[0][i]);
sorted[i] = tree[0][i];
}
sort(sorted + 1, sorted + 1 + n);
build(1, n, 0);
int a, b, c,d;
printf("Case #%d:\n", icase++);
while (k--)
{
scanf("%d%d%d%d", &a, &b, &c,&d);
printf("%d\n", gao(a, b, c,d));
}
}
}