Round A 2020 - Kick Start 2020 Workout【二分】

Problem

Tambourine has prepared a fitness program so that she can become more fit! The program is made of N sessions. During the i-th session, Tambourine will exercise for M_i minutes. The number of minutes she exercises in each session are strictly increasing.

 

The difficulty of her fitness program is equal to the maximum difference in the number of minutes between any two consecutive training sessions.

To make her program less difficult, Tambourine has decided to add up to K additional training sessions to her fitness program. She can add these sessions anywhere in her fitness program, and exercise any positive integer number of minutes in each of them. After the additional training session are added, the number of minutes she exercises in each session must still be strictly increasing. What is the minimum difficulty possible?

Input

The first line of the input gives the number of test cases, T. T test cases follow. Each test case begins with a line containing the two integers N and K. The second line contains N integers, the i-th of these is M_i, the number of minutes she will exercise in the i-th session.

Output

For each test case, output one line containing Case #x: y, where x is the test case number (starting from 1) and y is the minimum difficulty possible after up to K additional training sessions are added.

Limits

Time limit: 20 seconds per test set.
Memory limit: 1GB.
1 ≤ T ≤ 100.
For at most 10 test cases, 2 ≤ N ≤ 10⁵.
For all other test cases, 2 ≤ N ≤ 300.
1 ≤ M_i ≤ 10⁹.
M_i < M_i+1 for all i.

Test set 1

K = 1.

Test set 2

1 ≤ K ≤ 10⁵.

Samples

Input 1	Output 1
1 3 1 100 200 230	Case #1: 50

Input 2	Output 2
3 5 2 10 13 15 16 17 5 6 9 10 20 26 30 8 3 1 2 3 4 5 6 7 10	Case #1: 2 Case #2: 3 Case #3: 1

Sample #1

In Case #1: Tambourine can add up to one session. The added sessions are marked in bold: 100 150 200 230. The difficulty is now 50.

 

Sample #2

In Case #1: Tambourine can add up to two sessions. The added sessions are marked in bold: 10 11 13 15 16 17 18. The difficulty is now 2.

In Case #2: Tambourine can add up to six sessions. The added sessions are marked in bold: 9 10 12 14 16 18 20 23 26 29 30. The difficulty is now 3.

In Case #3: Tambourine can add up to three sessions. The added sessions are marked in bold: 1 2 3 4 5 6 7 8 9 10. The difficulty is now 1. Note that Tambourine only added two sessions.

 

• Note #1: Only Sample #1 is a valid input for Test set 1. Consequently, Sample #1 will be used as a sample test set for your submissions.
• Note #2: Unlike previous editions, in Kick Start 2020, all test sets are visible verdict test sets, meaning you receive instant feedback upon submission.

题意：数组有N个递增的数字，希望加K个数字进去使得数字间的差值最大的最小。

解法：输入的时候遍历一遍数组，求出当前差值最大多少，然后二分这个差值，判断的时候就模拟一下需要插几个数字进去，数字够用肯定可以成功。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
#include<algorithm>
#include<map>
#include<stack>
#include<queue>
using namespace std;
#define PI 3.14159
#define mm(a,b) memset(a,b,sizeof(a))
const int maxn = 1e5 + 10;
const int inf = 0x3f3f3f3f;
int gcd(int a,int b) {
    return b>0 ? gcd(b,a%b):a;
}
int arr[maxn];
int n,k;
bool check(int d)
{
    int valid=k;
    for(int i=1;i<n;i++)
    {
        if(arr[i]-arr[i-1]>d)
        {
            if((arr[i]-arr[i-1])%d==0)
            {
                valid=valid-(arr[i]-arr[i-1])/d+1;
            }
            else valid=valid-(arr[i]-arr[i-1])/d;
            
            if(valid<0)
            {
                return false;
            }
        }
    }
    if(valid>=0) return true;
    return false;
}
int Solve(int l,int r)
{
    int ans=-1;
    while(l<=r)
    {
        int mid=(l+r)/2;
        if(check(mid))
        {
            ans=mid;
            r=mid-1;
        }
        else
        {
            l=mid+1;
        }
    }
    return ans;
}

int main()
{
    int T;
    scanf("%d",&T);
    int ccase=1;
    while(T--)
    {
        scanf("%d %d",&n,&k);
        int diff=0;
        scanf("%d",&arr[0]);
        for(int i=1;i<n;i++)
        {
            scanf("%d",&arr[i]);
            diff=max(diff,arr[i]-arr[i-1]);
        }
        int ans=Solve(1,diff);
        printf("Case #%d: %d\n",ccase,ans);
        ccase++;
    }
    return 0;
}