You are given N positive integers, denoted as x0, x1 ... xN-1. Then give you some intervals [l, r]. For each interval, you need to find a number x to make as small as possible!
Input
The first line is an integer T (T <= 10), indicating the number of test cases. For each test case, an integer N (1 <= N <= 100,000) comes first. Then comes N positive integers x (1 <= x <= 1,000, 000,000) in the next line. Finally, comes an integer Q (1 <= Q <= 100,000), indicting there are Q queries. Each query consists of two integers l, r (0 <= l <= r < N), meaning the interval you should deal with.
Output
For the k-th test case, first output “Case #k:” in a separate line. Then output Q lines, each line is the minimum value of . Output a blank line after every test case.
Sample Input
2 5 3 6 2 2 4 2 1 4 0 2 2 7 7 2 0 1 1 1
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
using namespace std;
const int maxn = 200010;
int tree[20][maxn];
int sorted[maxn];
int toleft[20][maxn];
long long leftsum[20][maxn], lsum, rsum, sum[maxn];//所列出的东西只用在需要计算区间内比第k数大（小）的和时候才加
int lnum, rnum;//同上
void build(int l, int r, int dep) {
if (l == r)return;
int mid = (l + r) >> 1;
int same = mid - l + 1;
for (int i = l; i <= r; i++) {
if (tree[dep][i] < sorted[mid])
same--;
}
int lpos = l;
int rpos = mid + 1;
for (int i = l; i <= r; i++) {
int spot = 0;
if (tree[dep][i] < sorted[mid]) {
tree[dep + 1][lpos++] = tree[dep][i];
spot = 1;
}
else if (tree[dep][i] == sorted[mid] && same > 0) {
tree[dep + 1][lpos++] = tree[dep][i];
same--; spot = 1;
}
else
tree[dep + 1][rpos++] = tree[dep][i];
leftsum[dep][i] = leftsum[dep][i - 1] + spot * tree[dep][i];//不需要时去除
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 mid = (L + R) >> 1;
int cnt = toleft[dep][r] - toleft[dep][l - 1];
if (cnt >= k) {
int newl = L + toleft[dep][l - 1] - toleft[dep][L - 1];
int newr = newl + cnt - 1;
return query(L, mid, newl, newr, dep + 1, k);
}
else {
int newr = r + toleft[dep][R] - toleft[dep][r];
int newl = newr - (r - l - cnt);
lnum += cnt;//不需要时去除
lsum += leftsum[dep][r] - leftsum[dep][l - 1];//不需要时去除
return query(mid + 1, R, newl, newr, dep + 1, k - cnt);
}
}
int main() {
int n, m;
int te; scanf("%d", &te);
sum[0] = 0;
for (int i = 0; i < 20; i++)
leftsum[i][0] = 0;
int cas = 1;
while (te--) {
scanf("%d", &n);
for (int i = 1; i <= n; i++) {
scanf("%d", &tree[0][i]);
sorted[i] = tree[0][i];
sum[i] = sum[i - 1] + tree[0][i];
}
sort(sorted + 1, sorted + n + 1);
build(1, n, 0);
int s, t, k;
scanf("%d", &m);
printf("Case #%d:\n", cas++);
while (m--) {
scanf("%d%d", &s, &t);
s++, t++;//!
lsum = lnum = 0;
k = (t - s) / 2 + 1;//中位数的所在序号
int spot = query(1, n, s, t, 0, k); //中位数是多少
rnum = (t - s + 1 - lnum);//排序后比所求spot（包含spot这个点）右边的数的数量，lsum相反（不包括spot）
rsum = (sum[t] - sum[s - 1] - lsum);//排序后比所求spot（包含spot这个点）右边的数的和，rsum相反（不包括spot）
long long ans = rsum - lsum + spot * (lnum - rnum);
printf("%lld\n", ans);
}
printf("\n");
}
return 0;
}