（HDU 5783）Divide the Sequence <思维水题> 多校训练5_hdu 5783 divide the sequence-优快云博客

本文介绍了一个经典的编程竞赛问题——分割序列。目标是将一个整数序列分割成尽可能多的连续子序列，使得每个子序列的所有前缀和都不小于0。文章提供了一种有效的算法实现，并附带了AC代码。

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Divide the Sequence
Problem Description
Alice has a sequence A, She wants to split A into as much as possible continuous subsequences, satisfying that for each subsequence, every its prefix sum is not small than 0.

Input
The input consists of multiple test cases.
Each test case begin with an integer n in a single line.
The next line contains n integers A1,A2⋯An.
1≤n≤1e6
−10000≤A[i]≤10000
You can assume that there is at least one solution.

Output
For each test case, output an integer indicates the maximum number of sequence division.

Sample Input
6
1 2 3 4 5 6
4
1 2 -3 0
5
0 0 0 0 0

Sample Output
6
2
5

Author
ZSTU

Source
2016 Multi-University Training Contest 5

题意：
有一个大小为n的整型数组。现在要将他分成尽可能多的连续的子序列，满足每一子序列的所有前缀后大于等于0，问你最多可以分为多少个？

分析：
搞不清楚题目为什么要给2.5秒。。。
直接从后往前少一遍，若a[i]>=0 则将它单独分为一个子序列，若a[i]<0,则为使所有的前缀后>=0，那么我们要将以a[i]结尾的连续的一段和>=0的部分分为一个子序列即可。

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
using namespace std;

const int maxn = 1e6 + 10;
int a[maxn];

int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
        }
        int ans = 0;
        long long res = 0;
        for(int i=n;i>=1;)
        {
            res = res + a[i];
            if(res >= 0)
            {
                ans++;
                i--;
                res = 0;
            }
            else
            {
                while(res < 0)
                {
                    res = res + a[--i];
                }
                ans++;
                i--;
                res = 0;
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}