MYF

HDU 5783 Divide the Sequence

Posted on | In | | Visitors

题目链接

HDU 5783

题目类型：签到水题

题目来源：2016 Multi-University Training Contest 5

题目分析

题目大意

给出n个数，问这n个数最多可以组成多少个连续子区间，使每个子区间的前缀和不小于零

解析

首先先明确一个概念——前缀和。

前缀和是对于一个序列来讲，长度为n则有n个前缀和，第i个前缀和为区间内前i个数的累加。

从后往前遍历即可，找到一段大于零的就ans++即可。

代码

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
#include <set>
#include <map>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <string>
#include <vector>
#include <iomanip>
#include <bitset>
#include <cstring>
#include <iostream>
#include <deque>
#include <algorithm>
#define Memset(a,val) memset(a,val,sizeof(a))
#define PI acos(-1)
#define PB push_back
#define MP make_pair
#define rt(n)       (i == n ? '\n' : ' ')
#define hi         printf("Hi----------\n")
#define IN freopen("input.txt","r",stdin);
#define OUT freopen("output.txt","w",stdout);
#define debug(x) cout<<"Debug : ---"<<x<<"---"<<endl;
#define debug2(x,y) cout<<"Debug : ---"<<x<<" , "<<y<<"---"<<endl;
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
typedef pair<int,int> PII;
typedef long long ll;
const int maxn=1e6+5;
const int mod=1000000007;
const int INF=0x3f3f3f3f;
const double eps=1e-8;
ll a[maxn];
int main(){
    int n;
    while (~scanf("%d",&n)) {
        for (int i = 1; i<=n; i++) {
            scanf("%lld",a+i);
        }
        int ans =0;
        ll tmp = 0;
        for (int i=n;i>=1; i--) {
            tmp+=a[i];
            if (tmp>=0) {
                tmp=0;
                ans++;
            }
        }
        cout<<ans<<endl;
    }
}