Divide the Sequence

题目连接：

http://acm.hdu.edu.cn/showproblem.php?pid=5783

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

Hint

题意

说的是，给你n个数，让你分成最多的块，使得每一块的任何前缀，都是大于0的。

问你最多分成多少块

题解：

把长度为n的序列分成尽量多的连续段，使得每一段的每个前缀和都不小于0。保证有解。 从后往前贪心分段即可。

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e6+7;
int n;
long long a[maxn];
int main(){
    while(scanf("%d",&n)!=EOF){
        for(int i=1;i<=n;i++){
            scanf("%I64d",&a[i]);
        }
        long long flag = 0;
        long long ans = 0;
        long long now = 0;
        for(int i=n;i>=1;i--){
            if(flag==0&&a[i]>=0){
                ans++;
                flag = 0;
            }
            else if(flag==0&&a[i]<0){
                flag=1;
                now+=a[i];
                ans++;
            }else if(flag==1){
                now+=a[i];
                if(now>=0){
                    now=0,flag=0;
                }
            }
        }
        cout<<ans<<endl;
    }
}