2016 Al-Baath University Training Camp Contest-1 I. March Rain —— 二分

题目链接：http://codeforces.com/problemset/gymProblem/101028/I

I. March Rain

time limit per test

2 seconds

memory limit per test

64 megabytes

input

standard input

output

standard output

It is raining again! Youssef really forgot that there is a chance of rain in March, so he didn't fix the roof of his house. Youssef's roof is 1-D, and it contains n holes that make the water flow into the house, the position of hole i is denoted as xi where (0 ≤ i < n). Youssef has to put strips at the bottoms of those holes in order to prevent the water from flowing. Let's say there is a hole in position 4 and another hole in position 6, and Youssef decided to use a strip of length 3 to cover those two holes, then he places the strip from position 4 to 6 (it covers positions 4,5,6) and it covers the two holes. He can buy exactly k strips, and he must pay a price equal to the longest strip he buys. What is the minimum length l he can choose as the longest strip in order to keep his house safe?

Input

The input consists of several test cases. The first line of the input contains a single integer T, the number of the test cases. Each test case consists of two lines: the first line contains two space-separated integers, n and k (1 ≤ k < n ≤ 100000), denoting the number of the holes in the roof, and the number of the strips he can buy respectively. The second line of the test case contains n integers (x0, x1, ..., xn - 1): (0 ≤ xi ≤ 109), denoting the positions of holes (these numbers are given in an increasing order).

Output

For each test case print a single line containing a single integer denoting the minimum length l he can choose in order to buy k strips (the longest of them is of length l) and cover all the holes in his house using them.

Example

input

3
5 2
1 2 3 4 5
7 3
1 3 8 9 10 14 17
5 3
1 2 3 4 20

output

3
4
2

Note

In the second test case the roof looks like this before and after putting the strips.

题解：

一开始以为是区间覆盖问题。后来想到可以用二分来寻找答案。由于付款金额依照最长长条的长度。所以就把每条长条都想成是最长的。二分长度，然后判断当前长度是否能覆盖完所有漏洞。

代码如下：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
#include<map>
#include<string>
#include<set>
using namespace std;
#define pb push_back
#define ms(a, b) memset(a,b,sizeof(a));
typedef long long LL;
const int inf = 0x3f3f3f3f;
const int maxn = 250000;

int a[100005], T, n,k;

int test(int len)
{
    int now= 0, cnt = 0;
    for(int i = 0; i<n; i++)
    {
        if(a[i]>now)
        {
            cnt++;
            now = a[i]+len-1;
            if(cnt==k)//如果用完了k条时，判断是否已经覆盖完
            {
                if(now>=a[n-1])
                    return 1;
                else
                    return 0;
            }
            if(now>=a[n-1])//如果没用完k条，就覆盖完的话，肯定可以
                return 1;
        }
    }
    return 0;
}

int main()
{
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&k);
        for(int i = 0; i<n; i++)
        {
            scanf("%d",&a[i]);
        }

        int h = 1, t = a[n-1];
        while(t%k) t++;//注意这一步，将t自加到能整除k为止

        int mid;
        while(h<=t)
        {
            mid = (h+t)/2;
            if(test(mid))
                t = mid-1;
            else
                h = mid+1;
        }

        printf("%d
",h);
    }
    return 0;
}