HDU 5783 Divide the Sequence（水题）

Divide the Sequence

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 218    Accepted Submission(s): 130

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

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题意：给你一个序列，让你尽可能分成更多的子序列。使得每个子序列的前缀和都大于或等于0 。输出这个的子序列   的总个数。

题解：倒序遍历序列。如果遇到的是正数，那它就是一个符合条件的子序列，如果是负数，那就继续往前遍历，直到前缀和大于等于0，就是又一个符合条件的子序列。

AC代码：

#include<bits/stdc++.h>
long long a[1000010];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
    	long long sum=0;
        long long num=0;
        memset(a,0,sizeof(a));
        for(int i=1;i<=n;i++)
        {
            scanf("%I64d",&a[i]);
        }
        for(int i=n;i>=1;--i)
        {

            if(a[i]>=0)
            {
			    sum++; //子序列长度 
                a[i]=0;
            }
            else
            {
                num=a[i];
                a[i-1]+=num;

            }
        }
        printf("%I64d\n",sum);
    }
    return 0; 
}