Title Description
A sequence of length n is given, and the number of successive sub-sequences of all elements and in the [L,r] range is obtained.
Input
The first line contains three integers n,l and R, each representing the number of sushi trays, the lower and upper limits of satisfaction. The second line contains n integer AI, which represents the satisfaction of small z on sushi. n≤100000,| Ai|≤100000,0≤l, R≤10^9
Output
Only one row, which contains an integer, indicates how many choices are available to make the sum of the satisfaction of small z not less than L and not higher than R.
Sample input
5 5 9
1 2 3) 4 5
Sample output
6
Exercises
Discretization + Tree-like array
Subtract the sequence from the conversion prefix, which is the number of $x>y$ that satisfies the $l\le sum[x]-sum[y]\le R.
Then we enumerate the $x$, we can get the range of $y$, which requires the number of $y$ of the previous satisfying condition. You can maintain a tree array of 1 to the current location, query the number in the tree array, and then add that number to the tree array. Because of the large data range, discretization is required.
Time Complexity $o (n\log N) $
#include <cstdio> #include <algorithm> #define N 100010#define now v + 1, v + N + 2using namespace Std;typedef Long Long ll;ll sum[n], V[n];int f[n], n;inline void Add (int x) { int i; for (i = x; I <= n + 1; i + = i & i) f[i] + +;} inline int query (int x) { int i, ans = 0; for (i = x; i; i-= i & i) ans + = f[i]; return ans;} int main () { int i; ll L, r, ans = 0; scanf ("%d%lld%lld", &n, &l, &r); for (i = 1; I <= n; i + +) scanf ("%lld", &sum[i]), sum[i] + = Sum[i-1], v[i] = sum[i]; Sort (now); for (i = 0; I <= n; i + +) ans + = query (Upper_bound (now, Sum[i] – L)-V-1)-Query (Lower_bound (now, Sum[i]-R)- V-1), add (Lower_bound (now, sum[i])-V); printf ("%lld\n", ans); return 0;}
"bzoj4627" [BeiJing2016] rotary sushi discretization + tree-like array