poisoned dagger

题目描述

Monocarp is playing yet another computer game. In this game, his character has to kill a dragon. The battle with the dragon lasts 100^{500}100500 seconds, during which Monocarp attacks the dragon with a poisoned dagger. The ii -th attack is performed at the beginning of the a_iai -th second from the battle start. The dagger itself does not deal damage, but it applies a poison effect on the dragon, which deals 11 damage during each of the next kk seconds (starting with the same second when the dragon was stabbed by the dagger). However, if the dragon has already been poisoned, then the dagger updates the poison effect (i.e. cancels the current poison effect and applies a new one).

For example, suppose k = 4k=4 , and Monocarp stabs the dragon during the seconds 22 , 44 and 1010 . Then the poison effect is applied at the start of the 22 -nd second and deals 11 damage during the 22 -nd and 33 -rd seconds; then, at the beginning of the 44 -th second, the poison effect is reapplied, so it deals exactly 11 damage during the seconds 44 , 55 , 66 and 77 ; then, during the 1010 -th second, the poison effect is applied again, and it deals 11 damage during the seconds 1010 , 1111 , 1212 and 1313 . In total, the dragon receives 1010 damage.

Monocarp knows that the dragon has hh hit points, and if he deals at least hh damage to the dragon during the battle — he slays the dragon. Monocarp has not decided on the strength of the poison he will use during the battle, so he wants to find the minimum possible value of kk (the number of seconds the poison effect lasts) that is enough to deal at least hh damage to the dragon.

输入格式

The first line contains a single integer tt ( 1 \le t \le 10001≤t≤1000 ) — the number of test cases.

The first line of the test case contains two integers nn and hh ( 1 \le n \le 100; 1 \le h \le 10^{18}1≤n≤100;1≤h≤1018 ) — the number of Monocarp's attacks and the amount of damage that needs to be dealt.

The second line contains nn integers a_1a1 , a_2a2 , ..., a_nan ( 1 \le a_i \le 10^9; a_i < a_{i + 1}1≤ai≤109;ai<ai+1 ), where a_iai is the second when the ii -th attack is performed.

输出格式

For each test case, print a single integer — the minimum value of the parameter kk , such that Monocarp will cause at least hh damage to the dragon.

题意翻译

Monocarp 要斩杀一头巨龙，巨龙的血量为 h。Monocarp 可以发动 n 次攻击，每次攻击发动于时间 ai，持续时间 k 秒，每持续 1 秒就会让巨龙的血量 -1。如果两个攻击的间隔小于 k，后一个攻击发动时会停止前一个攻击的效果。

Monocarp 想知道斩杀这头巨龙的 k 的最小值是多少。

输入输出样例

输入 #1

4
2 5
1 5
3 10
2 4 10
5 3
1 2 4 5 7
4 1000
3 25 64 1337

输出 #1

说明/提示

In the first example, for k=3k=3 , damage is dealt in seconds [1, 2, 3, 5, 6, 7][1,2,3,5,6,7] .

In the second example, for k=4k=4 , damage is dealt in seconds [2, 3, 4, 5, 6, 7, 10, 11, 12, 13][2,3,4,5,6,7,10,11,12,13] .

In the third example, for k=1k=1 , damage is dealt in seconds [1, 2, 4, 5, 7][1,2,4,5,7] .

只给一周时间做5道二分题，本菜鸟是真的会“谢”。

拿到这道题，我是毫无思绪，于是上了vjudge看了别人的题解，哟嘿，很开心第一个题解就很短，

然鹅，看了半天，为什么要left<=right呀，为什么返回的是right+1,为什么另一个题解写的返回left呀........

好晕啊，即使我照着别人的题解敲一遍，还是搞不懂，然后看了那么多题解，里面好像有涉及什么二分模板，好吧，这肯定是个我不知道的套路，于是我就去查，终于在一本算法书上找到了它。

two days later.......

我终于理解了二分模板，好吧，让我再次面对它吧，果然，题解一下子就变得和蔼可亲了呢

I got it!!!!!!

先来分析一下这个题：这个思路借鉴了一位vjudge用户的题解

AC题解：

#include<cstdio>
#include<algorithm>
using namespace std;
#define ll long long
//用到long long 类型是因为防止在使用int时，mid=left+right 超过int最大数值
ll a[105],n,h;
int main(){int t;scanf("%d",&t);while(t--){scanf("%lld%lld",&n,&h);for(int i=1;i<=n;++i)scanf("%lld",&a[i]);ll left=1,right=1e18;//即使是2倍的right也不会超过long long的极限值 //二分核心步骤来了 while(left<right){ll mid=left+right>>1;//右移符>>1,相当于除以2（限于整数） ll ans=mid;for(int i=1;i<n;i++)ans+=min(mid,a[i+1]-a[i]);//ans的最终值即为通过k求出的总伤害 if(ans>=h)right=mid;elseleft=mid+1;}printf("%lld\n",left);}return 0;
}

我第一次接触那部分二分步骤模板的时候可是懵逼的很。

本题用到的二分模板：这是左右区间为闭区间的模板

//“寻找有序序列第一个满足某条件的元素的位置 ”问题的固定模板
//这个“某条件”一定是在序列中从左到右先不满足，然后满足的（否则把该条件取反即可）
//二分区间为左闭右闭的[left,right]
int solve(int left,int right){int mid;while(left<right){mid=(left+right)/2;if(条件成立){right=mid;}else{left=mid+1;}}return left;//循环结束的条件是left==right,则也可以返回right
}
/*如果想要寻找最后一个满足“条件C”的元素的位置,
可以先求第一个满足！C的位置，然后将该位置减1就行了 */

本题也可以用这个模板：左开右闭区间的模板

//解决“寻找有序序列第一个满足某条件的元素的位置”问题的固定模板
//这个“某条件”一定是在序列中从左到右先不满足，然后满足的（否则把该条件取反即可）
//二分区间为(left,right]，left比最小取值小1，才能覆盖所有取值
int solve(int left,int right){int mid;while(left+1<right){mid=(left+right)/2;if(条件成立)right=mid;elseleft=mid;}return right;//也可以返回left+1，但不能返回left
}

另外，解决“序列中是否存在满足某条件的元素”问题的模板如下：

//这里以在一个递增序列里查找是否存在等于x的元素为例
int solve(int a[],int left ,int right,int x){int mid;while(left<=right){mid=(left+right)/2;if(a[mid]==x)return mid;else if(a[mid]>x)right=mid-1;elseleft=mid+1;}return -1;
}

这里只是给出了三个模板，我是从胡凡，曾磊写的《算法笔记》里看的。有兴趣的朋友可以看书中原文：我拍了图片放在这个链接里面了

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