hdu 5783 水

lice 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 nn integers A1,A2⋯AnA1,A2⋯An.
1≤n≤1e61≤n≤1e6
−10000≤A[i]≤10000−10000≤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

题意
题意每一个前缀和为非负的串可以划分成一个区间
问最多可以划分多少个区间

做法
反着扫
扫到非负了就加起来独自成为一个区间即可
保证了至少有一解，就少了很多特殊情况。

#include <bits/stdc++.h>using namespace std;const int N = 1e6+100;int a[N];int main(){    int n;    while(scanf("%d",&n)!=EOF)    {        for(int i=0;i<n;i++)            scanf("%d",&a[i]);        long long sum=0;        int ans=0;        for(int i=n-1;i>=0;i--)        {            sum+=a[i];            if(sum>=0)            {                sum=0;                ans++;            }        }        cout<<ans<<endl;    }}