D. Credit Card
time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Recenlty Luba got a credit card and started to use it. Let's consider n consecutive days Luba uses the card.

She starts with 0 money on her account.

In the evening of i-th day a transaction ai occurs. If ai > 0, then ai bourles are deposited to Luba's account. If ai < 0, then ai bourles are withdrawn. And if ai = 0, then the amount of money on Luba's account is checked.

In the morning of any of n days Luba can go to the bank and deposit any positive integer amount of burles to her account. But there is a limitation: the amount of money on the account can never exceed d.

It can happen that the amount of money goes greater than d by some transaction in the evening. In this case answer will be «-1».

Luba must not exceed this limit, and also she wants that every day her account is checked (the days when ai = 0) the amount of money on her account is non-negative. It takes a lot of time to go to the bank, so Luba wants to know the minimum number of days she needs to deposit some money to her account (if it is possible to meet all the requirements). Help her!

Input

The first line contains two integers nd (1 ≤ n ≤ 105, 1 ≤ d ≤ 109) —the number of days and the money limitation.

The second line contains n integer numbers a1, a2, ... an ( - 104 ≤ ai ≤ 104), where ai represents the transaction in i-th day.

Output

Print -1 if Luba cannot deposit the money to her account in such a way that the requirements are met. Otherwise print the minimum number of days Luba has to deposit money.

Examples
input
5 10
-1 5 0 -5 3
output
0
input
3 4
-10 0 20
output
-1
input
5 10
-5 0 10 -11 0
output
2

【题意】:白天存钱,晚上加减,一旦见零,马上非负。钱有上限,最少几存。

【分析】:

【代码】:

Educational Codeforces Round 33 (Rated for Div. 2) D. Credit Card的更多相关文章

  1. Educational Codeforces Round 33 (Rated for Div. 2) E. Counting Arrays

    题目链接 题意:给你两个数x,yx,yx,y,让你构造一些长为yyy的数列,让这个数列的累乘为xxx,输出方案数. 思路:考虑对xxx进行质因数分解,设某个质因子PiP_iPi​的的幂为kkk,则这个 ...

  2. Educational Codeforces Round 33 (Rated for Div. 2) F. Subtree Minimum Query(主席树合并)

    题意 给定一棵 \(n\) 个点的带点权树,以 \(1\) 为根, \(m\) 次询问,每次询问给出两个值 \(p, k\) ,求以下值: \(p\) 的子树中距离 \(p \le k\) 的所有点权 ...

  3. Educational Codeforces Round 33 (Rated for Div. 2) 题解

    A.每个状态只有一种后续转移,判断每次转移是否都合法即可. #include <iostream> #include <cstdio> using namespace std; ...

  4. Educational Codeforces Round 33 (Rated for Div. 2)A-F

    总的来说这套题还是很不错的,让我对主席树有了更深的了解 A:水题,模拟即可 #include<bits/stdc++.h> #define fi first #define se seco ...

  5. Educational Codeforces Round 33 (Rated for Div. 2) C. Rumor【并查集+贪心/维护集合最小值】

    C. Rumor time limit per test 2 seconds memory limit per test 256 megabytes input standard input outp ...

  6. Educational Codeforces Round 33 (Rated for Div. 2) B. Beautiful Divisors【进制思维/打表】

    B. Beautiful Divisors time limit per test 2 seconds memory limit per test 256 megabytes input standa ...

  7. Educational Codeforces Round 33 (Rated for Div. 2) A. Chess For Three【模拟/逻辑推理】

    A. Chess For Three time limit per test 1 second memory limit per test 256 megabytes input standard i ...

  8. Educational Codeforces Round 33 (Rated for Div. 2)

    A. Chess For Three time limit per test 1 second memory limit per test 256 megabytes input standard i ...

  9. Educational Codeforces Round 33 (Rated for Div. 2) D题 【贪心:前缀和+后缀最值好题】

    D. Credit Card Recenlty Luba got a credit card and started to use it. Let's consider n consecutive d ...

随机推荐

  1. C#中接口与抽象类的区别

    接口与抽象类是面试中经常会考到的点,容易混淆.首先了解下两者的概念: 一.抽象类:      抽象类是特殊的类,只是不能被实例化:除此以外,具有类的其他特性:重要的是抽象类可以包括抽象方法,这是普通类 ...

  2. 剑指Offer - 九度1283 - 第一个只出现一次的字符

    剑指Offer - 九度1283 - 第一个只出现一次的字符2013-11-21 21:13 题目描述: 在一个字符串(1<=字符串长度<=10000,全部由大写字母组成)中找到第一个只出 ...

  3. Oracle 遇到的问题:dos命令下imp导入数据时出错

    赋予用户dba权限:很多情况下会遇到没有权限需要输入用户名及密码才能导入 --已知被赋予权限的用户名为:batch --第一步 登陆 sqlplus /nolog sql>conn /as sy ...

  4. (原)Unreal渲染模块 源码和实例分析说明

    @author:白袍小道 说明 1.由于小道就三境武夫而已,而UE渲染部分不仅管理挺大,而且牵扯技术和内容驳杂,所以才有这篇梳理. 2.尽量会按书籍和资料,源码,小模块的调试和搬山(就是敲键盘)..等 ...

  5. == 与 equals 之区别

    "=="和equals方法究竟有什么区别? (单独把一个东西说清楚,然后再说清楚另一个,这样,它们的区别自然就出来了,混在一起说,则很难说清楚) ==操作符专门用来比较两个变量的值 ...

  6. hnust 罚时计算器

    问题 F: 罚时计算器 时间限制: 1 Sec  内存限制: 128 MB提交: 229  解决: 63[提交][状态][讨论版] 题目描述 一般 ACM程序设计比赛都是五个小时.但是比赛结束时,DB ...

  7. 史林枫:开源HtmlAgilityPack公共小类库封装 - 网页采集(爬虫)辅助解析利器【附源码+可视化工具推荐】

    做开发的,可能都做过信息采集相关的程序,史林枫也经常做一些数据采集或某些网站的业务办理自动化操作软件. 获取目标网页的信息很简单,使用网络编程,利用HttpWebResponse.HttpWebReq ...

  8. 想玩API,这些套路我来告诉你!

    小伙伴是不是时常听说各种api接口的问题呢,可能许多人第一感觉:那是什么个玩意儿,那么多人回去研究它,今天思梦PHP小编就来为你揭开他的神秘的面纱,先看一下百度百科上面的官方的解释: 其实说白了就是为 ...

  9. IPv6的一些特殊地址

    IPv6的一些特殊地址   2008年7月3日第二次修正! 昨天是修正了地址部分,本想发上来的,没来得及.今天修正了NDP协议,接下来会是路由和转换部分. 总结一下各协议的精华:OSPF在于LSA,B ...

  10. [NOWCODER] myh的超级多项式

    题面 已知$f_i=(\sum_{j=1}^ka_j{v_j}^i )\bmod 1004535809$ 给定$v_1,v_2,\ldots,v_k,f_1,f_2,\ldots f_k$ 求$f_n ...