HDU 5783 Divide the Sequence

Divide the Sequence

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 649    Accepted Submission(s): 331

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

 

 

 

解析：要使所有前缀和不小于0，只需从后向前扫描整个序列a[]，如果a[i]非负，则a[i]是一个满足条件的子序列；否则，继续向前扫描，直到前缀和非负，形成一个满足条件的子序列。如此进行贪心即可。

 

 

 

#include <cstdio>
#define ll long long

const int MAXN = 1e6+5;
int a[MAXN];

int main()
{
    int n;
    while(~scanf("%d", &n)){
        for(int i = 0; i < n; ++i)
            scanf("%d", &a[i]);
        ll sum = 0;
        int ans = 0;
        for(int i = n-1; i >= 0; --i){
            sum += a[i];
            if(sum >= 0){
                ++ans;
                sum = 0;
            }
        }
        printf("%d
", ans);
    }
    return 0;
}