HDU 5783 Divide the Sequence

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本文介绍了一种算法，该算法用于将一个整数序列尽可能多地分割成连续子序列，确保每个子序列的所有前缀和都不小于0。文章提供了问题描述、输入输出格式及样例，并附带了实现代码。

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

Author

ZSTU

Source

2016 Multi-University Training Contest 5

Recommend

wange2014

题意：给你一个序列，让你尽可能分成更多的子序列。
使得每个子序列的前缀和都大于或等于0 。
输出这个的子序列的总个数。

注意前缀和的概念
不会有像 3 2 1 -99 100 这样的非法数据
因为有前缀和小于0

根据概念反过来扫

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>
#include <cmath>
#include <stack>
#include <string>
#include <map>
#include <set>
#define pi acos(-1)
#define LL long long
#define ULL unsigned long long
#define inf 0x3f3f3f3f
#define INF 1e18
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
using namespace std;
typedef pair<int, int> P;
const int maxn = 1e6 + 5;

LL a[maxn];
int main(void)
{
//	freopen("C:\\Users\\wave\\Desktop\\NULL.exe\\NULL\\in.txt","r", stdin);
    int n, i, j;
    LL sum, cnt;
    while (~scanf("%d", &n))
    {
        for (i = 1; i <= n; i++)
            scanf("%I64d", &a[i]); //用cin会T的
        cnt = sum = 0;
        for (i = n; i >= 1; i--){
            sum += a[i];
            if (sum >= 0){
                sum = 0;
                cnt++;
            }
        }
        printf("%I64d\n", cnt);
    }

	return 0;
}