Saruman's Army 贪心

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 ofR 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 integerR, the maximum effective range of all palantirs (where 0 ≤ R ≤ 1000), and an integern, the number of troops in Saruman’s army (where 1 ≤ n ≤ 1000). The next line contains n integers, indicating the positionsx1, …, xn of each troop (where 0 ≤ xi ≤ 1000). The end-of-file is marked by a test case withR = 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.

题目大意：白萨鲁曼必须带领他的军队沿着一条笔直的道路, 从艾辛格到头盔的深处。为了跟踪他的部队, 萨鲁曼在部队中分配看到了 palantirs 的石块。每个球有一个最大有效范围的 R 单位, 并必须在军队中携带的一些部队 (即, palantirs 不允许 "自由浮动" 在半空中)。帮助萨鲁曼通过确定萨鲁曼所需的最小数量来控制中土地球, 以确保他的每一个爪牙都在某些球的 R 单位内 palantirs。

 

 

 

 

 

#include
#include
using namespace std;
int main()
{
    int r,n;
    int k;
    int i,j;
    int a[1005];
    int sum;
    while(scanf("%d %d",&r,&n)!=EOF)
    {
        if(r==-1&&n==-1)
            break;
        for(i=0;i