hdu-5783 Divide the Sequence(贪心)

题目链接：

Divide the Sequence

Time Limit: 5000/2500 MS (Java/Others)    

Memory Limit: 65536/65536 K (Java/Others)

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

题意：

问把这个序列分成任何前缀和都非负的子序列最多能分成多少个;

思路:

贪心,求出前缀和,从后往前对一个和找到它前边最近的小于等于它的分段就好;

AC代码：

/************************************************
┆  ┏┓　　　┏┓ ┆   
┆┏┛┻━━━┛┻┓ ┆
┆┃　　　　　　　┃ ┆
┆┃　　　━　　　┃ ┆
┆┃　┳┛　┗┳　┃ ┆
┆┃　　　　　　　┃ ┆ 
┆┃　　　┻　　　┃ ┆
┆┗━┓　  　┏━┛ ┆
┆　　┃　  　┃　　┆　　　　　　
┆　　┃　  　┗━━━┓ ┆
┆　　┃　 AC代马 　　┣┓┆
┆　　┃　        　　┏┛┆
┆　　┗┓┓┏━┳┓┏┛ ┆
┆　　　┃┫┫　┃┫┫ ┆
┆　　　┗┻┛　┗┻┛ ┆      
************************************************ */ 
 
 
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <bits/stdc++.h>
#include <stack>
 
using namespace std;
 
#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
 
typedef  long long LL;
 
template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
    for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
    F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
    if(!p) { puts("0"); return; }
    while(p) stk[++ tp] = p%10, p/=10;
    while(tp) putchar(stk[tp--] + '0');
    putchar('
');
}
 
const LL mod=1e9+7;
const double PI=acos(-1.0);
const int inf=1e9;
const int N=1e6+10;
const int maxn=2e3+14;
const double eps=1e-8;

int a[N];
LL sum[N];

inline void solve()
{

}

int main()
{      
        int n;
        while(cin>>n)
        {
            For(i,1,n)read(a[i]),sum[i]=sum[i-1]+a[i];
            int ans=0,cur=n,i=n-1;
            LL cursum=sum[cur];
            while(cur)
            {
                if(sum[i]<=cursum)
                {
                    ans++;
                    cur=i;
                    cursum=sum[cur];
                }
                i--;
            }
            printf("%d
",ans);
        }
        return 0;
}