HDU 5783 Divide the Sequence （贪心）

Divide the Sequence

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=5783

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

Source

2016 Multi-University Training Contest 5

##题意：

将数串A分解成尽可能多的字串，使得每个字串的任一前缀和都不小于零.


##题解：

贪心.
从后往前遍历，每个和为零的地方都可以分割一次.
注意用longlong.


##代码：
``` cpp
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define LL long long
#define eps 1e-8
#define maxn 1010000
#define mod 100000007
#define inf 0x3f3f3f3f
#define IN freopen("in.txt","r",stdin);
using namespace std;

int n;

LL num[maxn];

int main(int argc, char const *argv[])

{

//IN;

//int t; cin >> t;
while(scanf("%d", &n) != EOF)
{
    for(int i=1; i<=n; i++)
        scanf("%I64d", &num[i]);

    LL cur = 0;
    int ans = 0;
    for(int i=n; i>=1; i--) {
        cur += num[i];
        if(cur >=0) {
            ans++;
            cur = 0;
        }
    }

    printf("%d\n", ans);
}

return 0;

}