Saruman's Army

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在中土世界，萨鲁曼必须使用帕兰提尔（一种有范围限制的监视石）来确保他的军队在从艾辛格到海姆深谷的行军中得到全面监控。每颗帕兰提尔由一名士兵携带，且覆盖范围有限。本篇探讨了如何在给定士兵位置和帕兰提尔覆盖范围的情况下，确定最少需要多少颗帕兰提尔以确保所有士兵都在监视范围内。

Saruman’s Army

Saruman the White must lead his army along a straight path from Isengard to Helm’s Deep. To keep track of his forces, Saruman distributes seeing stones, known as palantirs, among the troops. Each palantir has a maximum effective range of R units, and must be carried by some troop in the army (i.e., palantirs are not allowed to “free float” in mid-air). Help Saruman take control of Middle Earth by determining the minimum number of palantirs needed for Saruman to ensure that each of his minions is within R units of some palantir.

Input

The input test file will contain multiple cases. Each test case begins with a single line containing an integer R, the maximum effective range of all palantirs (where 0 ≤ R ≤ 1000), and an integer n, the number of troops in Saruman’s army (where 1 ≤ n ≤ 1000). The next line contains n integers, indicating the positions x1, …, xn of each troop (where 0 ≤ xi ≤ 1000). The end-of-file is marked by a test case with R = n = −1.

Output

For each test case, print a single integer indicating the minimum number of palantirs needed.

Sample Input

0 3
10 20 20
10 7
70 30 1 7 15 20 50
-1 -1

Sample Output

2
4

Hint

In the first test case, Saruman may place a palantir at positions 10 and 20. Here, note that a single palantir with range 0 can cover both of the troops at position 20.

In the second test case, Saruman can place palantirs at position 7 (covering troops at 1, 7, and 15), position 20 (covering positions 20 and 30), position 50, and position 70. Here, note that palantirs must be distributed among troops and are not allowed to “free float.” Thus, Saruman cannot place a palantir at position 60 to cover the troops at positions 50 and 70.

C++编写：

#include<iostream>
#include<algorithm>
using namespace std;
const int MAX_N=1000;
int Army[MAX_N+5];
int R,n;     //R是最大有效范围，n代表军队个数

void min_mark()
{
    sort(Army,Army+n);
    int i=0,min=0;
    while(i<n)
    {
        int s=Army[i++];
        while(i<n && Army[i]<=s+R)   i++;
        int p=Army[i-1];       //p是新被标记的点
        while(i<n && Army[i]<=p+R)    i++;
        min++;
    }
    cout<<min<<endl;
}

int main()
{
    ios::sync_with_stdio(false);
    while(cin>>R>>n && R!=-1 & n!=-1)
    {
        for(int i=0;i<n;i++)
            cin>>Army[i];
        min_mark();
    }
}