Present

合集 - 刷题集(97)
1.妖梦拼木棒2.[USACO1.2]回文平方数 Palindromic Squares3.[COCI2012-2013#1] LJUBOMORA4.火柴棒等式5.Kieszonkowe6.[COCI2014-2015#5] TRAKTOR7.硬币翻转8.[蓝桥杯 2020 省 AB1] 走方格9.P2097 资料分发10.[AHOI2005]约数研究11.[NOIP2004 普及组] 火星人12.我焯,原神(这真是个题)13.滑雪 OJ365114.CV的人有福啦!YTU贪心训练2(部分注释)15.砝码称重16.警钟长鸣-----找零钱17.[USACO09OCT]Barn Echoes G18.【CSGRound2】光骓者的荣耀19.初见蓝桥杯20.[蓝桥杯 2022 省 B] 扫雷21.开餐馆22.P9241 [蓝桥杯 2023 省 B] 飞机降落23.P7083 [NWRRC2013] Energy Tycoon24.高精度25.P7954 [COCI2014-2015#6] PAPRIKA26.(输出路径搜索)[USACO06OCT] Cows on Skates G27.畜栏保留问题28.(贪心+搜索+剪枝)P8801 [蓝桥杯 2022 国 B] 最大数字29.蓝桥杯 入门----普及-30.[NOIP2016 提高组] 玩具谜题31.[NOIP2002 提高组] 字串变换(无向图查重)32.[NOIP2000 提高组] 单词接龙33.POJ 硬币34.动态规划学习 1(最长上升子序列问题)35.动态规划学习 2(背包问题及其变形)36. 切割回文37.[NOIP2001 提高组] 数的划分(剪枝)38.字符串转换数字,sscanf和sprintf大法39.[CTSC1997] 选课(树状DP)40. k 倍区间(同余定理,组合数)41.[蓝桥杯 2018 省 AB] 全球变暖42.马的遍历(对bfs的更深一层探讨)43.找零钱44.Pasha Maximizes45.Newspaper Headline46.Balanced Ternary String47.(数论)素数,质数48.(数论) 约数49.Helping the Nature50.(博弈论)Even Number Addicts
51.Present
52.Mr. Kitayuta's Colorful Graph(二维并查集,弱化版)53.Edgy Trees(dfs,并查集,快速幂,树形结构,红黑树)54.并查集55.再谈 前缀和,差分,离散化56.Best Cow Fences(前缀和+特殊二分)57.二分58.同类型,类背包动态规划,选地dp59.Primes on Interval(欧拉筛+二分+滑动窗口)60.字符串(KMP , Trie树 , STL)61.Codeforces1 #879 div.262.Add Modulo 10(数论,思维,数学,规律)63.期末考试YTU4035: Shmily(数学,等差数列)64.[POI2006] OKR-Periods of Words65.Getting Zero(Bfs)66.快速幂67.Dreaming of Freedom(数论,贪心)68.Scoring Subsequences69.Dolce Vita70.Cake Assembly Line71.Constructive Problem72.Forever Winter73.Maximum Sum(前缀和)74.Contrast Value(数学规律)75.Lamps(STL+双端队列)76.Long Long77.Sum in Binary Tree78.Apple Tree(树状搜索,树形DP)79.Strong Password(贪心思想)80.Come Together81.Notepad82.Hamiltonian Wall83.7.17作业84.逛画展(双指针)85.#88586.树状数组87.Subsequence Addition88.7.22作业89.Diverse Substrings90.Tracking Segments(二分,区间前缀)91.最大食物链计数92.线段树193.[TJOI2007] 线段(状态机)94.#87895.测试196.李白打酒97.砍竹子
收起
C. Present
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

Little beaver is a beginner programmer, so informatics is his favorite subject. Soon his informatics teacher is going to have a birthday and the beaver has decided to prepare a present for her. He planted n flowers in a row on his windowsill and started waiting for them to grow. However, after some time the beaver noticed that the flowers stopped growing. The beaver thinks it is bad manners to present little flowers. So he decided to come up with some solutions.

There are m days left to the birthday. The height of the i-th flower (assume that the flowers in the row are numbered from 1 to n from left to right) is equal to ai at the moment. At each of the remaining m days the beaver can take a special watering and water w contiguous flowers (he can do that only once at a day). At that each watered flower grows by one height unit on that day. The beaver wants the height of the smallest flower be as large as possible in the end. What maximum height of the smallest flower can he get?

Input

The first line contains space-separated integers n, m and w (1 ≤ w ≤ n ≤ 105; 1 ≤ m ≤ 105). The second line contains space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 109).

Output

Print a single integer — the maximum final height of the smallest flower.

Examples
input Copy
6 2 3
2 2 2 2 1 1
output Copy
2
input Copy
2 5 1
5 8
output Copy
9
Note

In the first sample beaver can water the last 3 flowers at the first day. On the next day he may not to water flowers at all. In the end he will get the following heights: [2, 2, 2, 3, 2, 2]. The smallest flower has height equal to 2. It's impossible to get height 3 in this test.

二分+差分

复制代码
//求最大的最小值问题,首先就要想到二分,接着看题,又是加减连续的一段区间
//加减连续的区间就要想到用差分数组来维护
//对于一组花,设x是最大的最小高度,那么每次二分x,有:
//枚举每次mid,即check函数: 如果当前值小于最大的最小值,那就要浇花,一直浇到最大的最小值,并记录下天数
//如果天数够,那证明这个方案可行,然后我们的答案再向后推推,看看有没有更大的答案
//如果不够,那就是这个答案过于大了,那就向前推推.
//浇花时要利用差分数组来进行操作
#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int N=1e5+10;
ll res,a[N],c[N],l=2e9,r=-2e9,n,m,w;
bool check(int h)
{
    ll day=0,now=0;//day代表把花浇到指定高度需要的天数
    for(int i=1;i<=n;i++){
        now+=c[i];//当前高度;
        if(now<h){
            day+=h-now;
            c[i]+=h-now;
            if(i+w<=n) c[i+w]-=h-now;
            now+=h-now;
        }
    }
    return day<=m;
}
int main()
{
    cin>>n>>m>>w;
    for(int i=1;i<=n;i++){
        cin>>a[i];
        l=min(l,a[i]);
        r=max(r,a[i]);
        c[i]=a[i]-a[i-1];//建立差分数组
    }
    r+=m+1;//因为要浇m天,右端点必须是最大的
    while(l<=r){
        ll mid=l+r>>1;
        if(check(mid)) res=mid,l=mid+1;
        else r=mid-1;
        for(int i=1;i<=n;i++) c[i]=a[i]-a[i-1];//每次都要重置差分数组
    }
    cout<<res<<endl;
    return 0;
}
复制代码

 


__EOF__

本文作者：Sakurajimamai
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