Poj 3618 registration

Token

Time Limit:1000 MS		Memory Limit:65536 K
Total Submissions:3826		Accepted:1850

Description

Bessie is traveling on a road teeming with interesting landmarks. The road is labeled just like a number line, and Bessie starts at the "origin "(X= 0). A totalN(1 ≤NLess than or equal to 50,000) landmarks are located at pointsX1,X2 ,...,XN(-100,000 ≤Xi≤ 100,000). Bessie wants to visit as your landmarks as possible before sundown, which occurs inT(1 ≤T≤ 1,000,000,000) minutes. She travels 1 distance unit in 1 minute.

Bessie will visit the landmarks in a particle order. since the landmarks closer to the origin are more important to Farmer John, she always heads for the unvisited landmark closest to the origin. no two landmarks will be the same distance away from the origin.

Help Bessie determine the maximum number of landmarks she can visit before the day ends.

Input

* Line 1: Two space-separated integers:TAndN
* Lines 2 ..N+ 1: LineI+ 1 contains a single integer that is the location ofITh landmark:Xi

Output

* Line 1: The maximum number of landmarks Bessie can visit.

Sample Input

25 510-38-71

Sample Output

4

QUESTION: Let's give you a bunch of points. Starting from the origin, the rule of the point is to give priority to the point closest to the origin. Ask the maximum number of points that can be reached in t seconds;

# Include <iostream> # include <algorithm> using namespace std; # define MAX 50000 + 5 struct point {int landmark, distance;} p [MAX]; bool cmp (point, point B) {if (. distance <B. distance) return true; return false;} int main () {int t, n; while (cin >>t> n) {for (int I = 0; I <n; I ++) {cin> p [I]. landmark; p [I]. distance = p [I]. landmark> 0? P [I]. landmark :(-p [I]. landmark);} sort (p, p + n, cmp);/* Testfor (int I = 0; I <n; I ++) cout <p [I]. landmark <"<p [I]. distance <endl; Test */int last = 0, I; for (I = 0; I <n; I ++) {int d = p [I]. landmark-last> 0? (P [I]. landmark-last) :( last-p [I]. landmark); if (t> = d) {t-= d; last = p [I]. landmark;} elsebreak;} cout <I <endl;} return 0 ;}

Poj 3618 registration