Divide the Sequence

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 70    Accepted Submission(s): 41

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

61 2 3 4 5 641 2 -3 050 0 0 0 0

 

Sample Output

625
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。
#include<stdio.h>#include<iostream>using namespace std;#include<string.h>#define maxn 10000long long a[1000020];int main(){    int n,m,k;    while(scanf("%d",&n)!=-1)    {        for(int i=0; i<n; i++)            scanf("%lld",&a[i]);        int sum=0;        for(int i=n-1; i>=0; i--)        {            if(a[i]>=0)            {                sum++;            }            else if(a[i]<0)            {                a[i-1]=a[i-1]+a[i];            }        }        printf("%d\n",sum);    }}