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
 
 
 
解析：要使所有字首和不小於0，只需從後向前掃描整個序列a[]，如果a[i]非負，則a[i]是一個滿足條件的子序列；否則，繼續向前掃描，直到字首和非負，形成一個滿足條件的子序列。如此進行貪心即可。
 
 
 
#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\n", ans);

    }

    return 0;

}