HDU5783（2016多校第五场）——Divide the Sequence（水水水）

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最新推荐文章于 2025-08-05 22:18:18 发布

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分类专栏：算法文章标签：算法 HDU

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本文介绍了一个算法问题，目标是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零。提供了问题描述、输入输出样例及解析思路。

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Divide the Sequence

Time Limit: 5000/2500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 962 Accepted Submission(s): 471

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

水题水题，就是求最多能把一个数组分成几份，保证每份都连续而且每个前缀和都大于0。

思路：从后往前扫，每遇到一个负数就继续往前知道可以使得前缀和大于0。

看代码把。

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
using namespace std;

const int INF=1000000007;
const int MAXN =1000010;
const int MAXM=1000010;

long long a[MAXN];

int main(){
    int n;
    while(scanf("%d",&n)!=EOF){
        for(int i=1;i<=n;i++)
            scanf("%I64d",a+i);
        int cnt=0;
        long long s=0;
        for(int i=n;i>=1;i--){
            if(a[i]>=0&&s==0)
                cnt++;
            else if(a[i]<0){
                s+=a[i];
            }
            else if(a[i]>=0&&s!=0){
                s+=a[i];
                if(s>=0)
                {
                    cnt++;
                    s=0;
                }
            }
        }
        printf("%d\n",cnt);
    }
}