Manao is solving a problem with the following statement:
He came up with a solution that produces the correct answers but is too slow. You are given the pseudocode of his solution, where the function getAnswer calculates the answer to the problem:
getAnswer(a[1..n], b[1..len], h)
answer = 0
for i = 1 to n-len+1
answer = answer + f(a[i..i+len-1], b, h, 1)
return answer
f(s[1..len], b[1..len], h, index)
if index = len+1 then
return 1
for i = 1 to len
if s[index] + b[i] >= h
mem = b[i]
b[i] = 0
res = f(s, b, h, index + 1)
b[i] = mem
if res > 0
return 1
return 0
Your task is to help Manao optimize his algorithm.
Input
The first line contains space-separated integers n, len and h (1 ≤ len ≤ n ≤ 150000; 1 ≤ h ≤ 109). The second line contains len space-separated integers b 1, b 2, …, b len (1 ≤ b i ≤ 109). The third line contains n space-separated integers a 1, a 2, …, a n (1 ≤ a i ≤ 109).
Output
Print a single number — the answer to Manao’s problem.
Examples
input
5 2 10
5 3
1 8 5 5 7
output
2
Solution:
#include <vector> #include <list> #include <map> #include <set> #include <deque> #include <stack> #include <bitset> #include <algorithm> #include <functional> #include <numeric> #include <utility> #include <sstream> #include <iostream> #include <iomanip> #include <cstdio> #include <cmath> #include <cstdlib> #include <ctime> #include <memory.h> using namespace std; const int N = 2000010; int mx[N], add[N], a[N], b[N]; void modify(int x, int l, int r, int ll, int rr, int v) { if (l > rr || ll > r) return; if (l >= ll && r <= rr) { add[x] += v; return; } if (add[x] != 0) { add[x + x] += add[x]; add[x + x + 1] += add[x]; add[x] = 0; } int y = (l + r) >> 1; modify(x + x, l, y, ll, rr, v); modify(x + x + 1, y + 1, r, ll, rr, v); mx[x] = mx[x + x + 1] + add[x + x + 1]; if (mx[x + x] + add[x + x] > mx[x]) mx[x] = mx[x + x] + add[x + x]; } int main() { int n, len, h; scanf("%d %d %d", &n, &len, &h); for (int i=1;i<=len;i++) scanf("%d", b+i); sort(b + 1, b + len + 1); reverse(b + 1, b + len + 1); for (int i=1;i<=n;i++) { int foo; scanf("%d", &foo); int ll = 0, rr = len; while (ll < rr) { int mid = (ll + rr + 1) >> 1; if (foo + b[mid] >= h) ll = mid; else rr = mid - 1; } a[i] = ll; } memset(mx, 0, sizeof(mx)); memset(add, 0, sizeof(add)); for (int i=1;i<=len;i++) modify(1, 0, len, i, i, -i); for (int i=1;i<=len-1;i++) modify(1, 0, len, a[i], len, 1); int ans = 0; for (int i=len;i<=n;i++) { modify(1, 0, len, a[i], len, 1); if (mx[1] + add[1] == 0) ans++; modify(1, 0, len, a[i - len + 1], len, -1); } printf("%d\n", ans); return 0; }
Related posts:
King Escape
Floyd-Warshall - finding all shortest paths
Optimal Subsequences (Hard Version)
Finding the rank of a matrix
Strange Function
Make It One
R3D3’s Summer Adventure
Suffix Tree. Ukkonen's Algorithm
New Year and Ancient Prophecy
Lazy Security Guard
New Year and the Acquaintance Estimation
GCD Table
Substring Search
Sereja and the Arrangement of Numbers
Interesting Subarray
Aroma's Search
Kuroni and Simple Strings
Number Transformation
Finding the Eulerian path in $O(M)$
Om Nom and Dark Park
Elections
M-numbers
Union of Doubly Linked Lists
JYPnation
To Hack or not to Hack
Permutations
Almost Same Distance
New Year Shopping
AND Graph
Earth Wind and Fire
Maximum Flow
Kirchhoff's theorem. Finding the number of spanning trees