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.

6

2

5

/*就是给你一个序列，让你求最多能分成多少段连续子序列，并且每段子序列的前缀个都得是大于等于0的；

思路就是从后往前遍历，要想分的最多那么有整数直接，当成一个数列就行了，有负数往前递加直到和为零。这里特别注意，数据要用long long要不然会溢出的

*/

#include <iostream>

#include <algorithm>

#include <stdio.h>

#include <string.h>

#define N 1000005

#define M (1<<15)+5

#define INF 0x3f3f3f

using namespace std;

long long n,a[N],sum;

int main()

{

    //freopen("in.txt","r",stdin);

    while(scanf("%lld",&n)!=EOF)

    {

        memset(a,,sizeof(a));

        for(int i=;i<n;i++)

        {

            scanf("%lld",&a[i]);

        }

        long long cur=;

        for(int i=n-;i>=;i--)//这个是首位枚举

        {

            if(a[i]>=)

                cur++;

            else

            {

                sum=;

                for(;i>=;i--)

                {

                    sum+=a[i];

                    if(sum>=)

                    {

                        cur++;

                        break;

                    }

                }

            }

        }

            printf("%lld\n",cur);

    }

    return ;

}