hdoj5783Divide the Sequence_hdoj5783Divide

Divide the Sequence Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 501    Accepted Submission(s): 275


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  

/* ***********************************************Author       : rycCreated Time : 2016-08-03 WednesdayFile Name    : E:\acmcode\hdoj\5783.cppLANGUAGE     : c++Copyright 2016 ryc All Rights Reserved************************************************ */#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<stack>#include<vector>#include<map>using namespace std;const int maxn=1000010;long long num[maxn];long long sum[maxn];long long pos[maxn];bool sign[maxn];int main(){    int n;    while(scanf("%d",&n)!=EOF){        for(int i=1;i<=n;++i){            scanf("%lld",&num[i]);            sum[i]=sum[i-1]+num[i];        }        memset(sign,false,sizeof(sign));        long long ans=0,temp=0,l=1;pos[0]=-1;        for(int i=1;i<=n;++i){            temp+=num[i];            while(temp<0&&l>0){                l=pos[l-1];                temp=sum[i]-sum[l-1];sign[l]=false;            }            sign[l]=true;pos[i]=l;l=i+1;temp=0;        }        for(int i=1;i<=n;++i){            if(sign[i])ans++;        }        printf("%lld\n",ans);    }    return 0;}