HDU_5783_DivideTheSequence（贪心）

Divide the Sequence

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 637    Accepted Submission(s): 330

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

 

Author

ZSTU

 

Source

2016 Multi-University Training Contest 5

 

Recommend

wange2014

题意

题意每一个前缀和为非负的串可以划分成一个区间

问最多可以划分多少个区间

做法

反着扫

扫到非负了就加起来独自成为一个区间即可

保证了至少有一解，就少了很多特殊情况。

#include <iostream>#include <stdio.h>#include <string.h>using namespace std;typedef long long LL;const int M=1e6+5;int nu[M];int main(){    int n;    while(scanf("%d",&n)!=EOF)    {        for(int i=1;i<=n;i++)            scanf("%d",&nu[i]);        int ans=0;        LL sum=0;        for(int i=n;i>=1;i--)        {            sum+=nu[i];            if(sum>=0)            {                ans++;                sum=0;            }            else                continue;        }        printf("%d\n",ans);    }    return 0;}