Divide the Sequence

题意:

给你一个序列A,问你最多能够分成多少个连续子序列,使得每个子序列的所有前缀和均不小于0

题解:

这题是比赛时候的水题,但我比的时候也就做出这一题, = =

首先我想的是把他们前缀和求出来,之后试了下样例,一点鸟用都没有,那正着不行就倒着试下呗,之后发现这样有用,我先想到的思路是求后缀和,只要>=0就ans++,但那时队友举出了反例,比如2 1 -3 3的时候应该是2,我的算法就是4,那接着马上就能想到如果后缀和大于0,那么就要把他赋为0,之后举了几组例子,没问题,写了几分钟,之后1A。

代码:

#include <bits/stdc++.h>
using namespace std; typedef long long ll;
const int INF=0x3f3f3f3f;
const ll LINF=0x3f3f3f3f3f3f3f3f;
#define PI(A) cout<<A<<endl
#define SI(N) cin>>N
#define SII(N,M) cin>>N>>M
#define cle(a,val) memset(a,(val),sizeof(a))
#define rep(i,b) for(int i=0;i<(b);i++)
#define Rep(i,a,b) for(int i=(a);i<=(b);i++)
#define reRep(i,a,b) for(int i=(a);i>=(b);i--)
#define dbg(x) cout <<#x<<" = "<<x<<endl
#define PIar(a,n) rep(i,n)cout<<a[i]<<" ";cout<<endl;
#define PIarr(a,n,m) rep(aa,n){rep(bb, m)cout<<a[aa][bb]<<" ";cout<<endl;}
const double EPS= 1e-9 ; /* ///////////////////////// C o d i n g S p a c e ///////////////////////// */ const int MAXN= 1000000 + 9 ; ll a[MAXN];
int N; int main()
{
iostream::sync_with_stdio(false);
cin.tie(0),cout.tie(0);
while(SI(N))
{
ll ans=0;
rep(i,N) SI(a[i]);
ll sum=0;
reRep(i,N-1,0)
{
sum+=a[i];
if (sum>=0) sum=0;
if (sum==0) ans++;
}
PI(ans);
}
return 0;
}

2016 Multi-University Training Contest 5 Divide the Sequence的更多相关文章

  1. 2016 Al-Baath University Training Camp Contest-1

    2016 Al-Baath University Training Camp Contest-1 A题:http://codeforces.com/gym/101028/problem/A 题意:比赛 ...

  2. 2015 Multi-University Training Contest 1 OO’s Sequence

    OO’s Sequence Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)T ...

  3. 2016 Al-Baath University Training Camp Contest-1 E

    Description ACM-SCPC-2017 is approaching every university is trying to do its best in order to be th ...

  4. 2016 Al-Baath University Training Camp Contest-1 A

    Description Tourist likes competitive programming and he has his own Codeforces account. He particip ...

  5. 2016 Al-Baath University Training Camp Contest-1 J

    Description X is fighting beasts in the forest, in order to have a better chance to survive he's gon ...

  6. 2016 Al-Baath University Training Camp Contest-1 I

    Description It is raining again! Youssef really forgot that there is a chance of rain in March, so h ...

  7. 2016 Al-Baath University Training Camp Contest-1 H

     Description You've possibly heard about 'The Endless River'. However, if not, we are introducing it ...

  8. 2016 Al-Baath University Training Camp Contest-1 G

    Description The forces of evil are about to disappear since our hero is now on top on the tower of e ...

  9. 2016 Al-Baath University Training Camp Contest-1 F

    Description Zaid has two words, a of length between 4 and 1000 and b of length 4 exactly. The word a ...

随机推荐

  1. linux网络故障解决方法

    一.检测工具 tcpdump:dump the traffic on a network,根据使用者的定义对网络上的数据包进行截获的包分析工具. tcpdump可以将网络中传送的数据包的“头”完全截获 ...

  2. 异步处理工具类:AsyncTask

    (一) AsyncTask,是android提供的轻量级的异步类.可以直接继承AsyncTask,在类中实现异步操作,可以通过接口实现UI进度更新,最后反馈执行的结果给UI主线程 .之所以有Handl ...

  3. php mcrypt 完全安装

    今天安装完 PHP ,访问某个功能时,  /var/log/httpd/error_log  中报如下错误: PHP Fatal error:  Call to undefined function ...

  4. Articles Every Programmer Must Read

    http://javarevisited.blogspot.sg/2014/05/10-articles-every-programmer-must-read.html Being a Java pr ...

  5. lucene 专业名词作用整理

    是否切词:对关键词是否切分,举例,姓名域的一个值:"张三" , 是否切分成"张"."三"等等多个term. 是否索引:建立索引的时候是否对该 ...

  6. Linux进程间通信-匿名管道

    前面我们讲了进程间通信的一种方式,共享内存.下面看一看另一种机制,匿名管道.1.什么是管道管道是一个进程的数据流到另一个进程的通道,即一个进程的数据输出作为另一个进程的数据输入,管道起到了桥梁的作用. ...

  7. Visual Studio 文件没发布出来

    解决办法是选择文件打开属性窗口找到生成操作,选项选择"内容",重新发布,OK,问题解决.

  8. JavaScript 全选函数的实现

    Html代码: <table id="purchase-info" class="table table-bordered table-hover table-st ...

  9. unity shader random number

    http://gamedev.stackexchange.com/questions/32681/random-number-hlsl

  10. 【转】关于B/S架构应用程序的权限设置分析和总结

    来自:http://www.cnblogs.com/zhouxunyu/p/3790122.html 分析:不同的用户登录到系统后赋予不同的操作权限,而用户存在于数据库中,标识用户权限的字段也保存在数 ...