Minimum Sum
Time Limit: 16000/8000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4611 Accepted Submission(s): 1046
Problem Description
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!
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.
. 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
Sample Output
Case #1:
6
4
Case #2:
0
0
Author
standy
裸题我都不会 我废了
现在还是懵逼状态
ans= 【L，R 】 （中位数左边的值-中位数右边的值 ） 如果 注意一下这个区间元素的个数
用一个sum数组累加
当进入左子树查找时ans+=查询区间进入右子树的元素和
当进入右子树查询时ans-=查询区间进入左子树的元素和。
1 #include <cstdio> 2 #include <algorithm> 3 #include <cstring> 4 #include <cmath> 5 using namespace std; 6 typedef long long LL; 7 const int maxn = 1e5 + 10; 8 int sorted[maxn]; 9 int num[20][maxn], val[20][maxn]; 10 LL sum[20][maxn]; 11 12 void build(int l, int r, int dep) { 13 if (l == r) { 14 sum[dep][l] = sum[dep][l - 1] + val[dep][l]; 15 return ; 16 } 17 int mid = (l + r) >> 1, same = mid - l + 1; 18 for (int i = l ; i <= r ; i++) { 19 if (val[dep][i] < sorted[mid]) same--; 20 sum[dep][i] += sum[dep][i - 1] + val[dep][i]; 21 } 22 int lpos = l, rpos = mid + 1; 23 for (int i = l ; i <= r ; i++) { 24 if (val[dep][i] < sorted[mid]) val[dep + 1][lpos++] = val[dep][i]; 25 else if (val[dep][i] == sorted[mid] && same > 0) { 26 val[dep + 1][lpos++] = val[dep][i]; 27 same--; 28 } else val[dep + 1][rpos++] = val[dep][i]; 29 num[dep][i] = num[dep][l - 1] + lpos - l; 30 } 31 build(l, mid, dep + 1) ; 32 build(mid + 1, r, dep + 1); 33 } 34 LL ans; 35 36 int query(int L, int R, int l, int r, int dep, int k) { 37 if (l == r) return val[dep][l]; 38 int mid = (L + R) >> 1; 39 int cnt = num[dep][r] - num[dep][l - 1]; 40 if (cnt >= k) { 41 int ee = r - L + 1 - (num[dep][r] - num[dep][L - 1]) + mid; 42 int ss = l - L - (num[dep][l - 1] - num[dep][L - 1]) + mid; 43 ans += sum[dep + 1][ee] - sum[dep + 1][ss]; 44 int newl = L + num[dep][l - 1] - num[dep][L - 1]; 45 int newr = newl + cnt - 1; 46 return query(L, mid, newl, newr, dep + 1, k); 47 } else { 48 int s = L + num[dep][l - 1] - num[dep][L - 1]; 49 int e = s + cnt - 1; 50 ans -= sum[dep + 1][e] - sum[dep + 1][s - 1]; 51 int newr = r + num[dep][R] - num[dep][r]; 52 int newl = newr - (r - l + 1 - cnt) + 1; 53 return query(mid + 1, R, newl, newr, dep + 1, k - cnt); 54 } 55 } 56 57 int main() { 58 int t, n, cas = 1, m, l, r; 59 scanf("%d", &t); 60 while(t--) { 61 scanf("%d", &n); 62 memset(val, 0, sizeof(val)); 63 memset(sum, 0, sizeof(sum)); 64 for (int i = 1 ; i <= n ; i++) { 65 scanf("%d", &val[0][i]); 66 sorted[i] = val[0][i]; 67 } 68 sort(sorted + 1, sorted + n + 1); 69 build(1, n, 0); 70 printf("Case #%d: ", cas++); 71 scanf("%d", &m); 72 while(m--) { 73 scanf("%d%d", &l, &r); 74 ans = 0; 75 l++, r++; 76 int temp = query(1, n, l, r, 0, (l + r) / 2 - l + 1); 77 if ((l + r) % 2) ans -= temp; 78 printf("%lld ", ans); 79 } 80 printf(" "); 81 } 82 return 0; 83 }