(HDU 5783)Divide the Sequence <思维水题> 多校训练5
2017-01-17 21:02
453 查看
Divide the Sequence
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
题意:
有一个大小为n的整型数组。现在要将他分成尽可能多的连续的子序列,满足每一子序列的所有前缀后大于等于0,问你最多可以分为多少个?
分析:
搞不清楚题目为什么要给2.5秒。。。
直接从后往前少一遍,若a[i]>=0 则将它单独分为一个子序列,若a[i]<0,则为使所有的前缀后>=0,那么我们要将以a[i]结尾的连续的一段和>=0的部分分为一个子序列即可。
AC代码:
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
题意:
有一个大小为n的整型数组。现在要将他分成尽可能多的连续的子序列,满足每一子序列的所有前缀后大于等于0,问你最多可以分为多少个?
分析:
搞不清楚题目为什么要给2.5秒。。。
直接从后往前少一遍,若a[i]>=0 则将它单独分为一个子序列,若a[i]<0,则为使所有的前缀后>=0,那么我们要将以a[i]结尾的连续的一段和>=0的部分分为一个子序列即可。
AC代码:
#include <iostream> #include <cstdio> #include <cstring> #include <string> #include <algorithm> #include <cmath> using namespace std; const int maxn = 1e6 + 10; int a[maxn]; int main() { int n; while(scanf("%d",&n)!=EOF) { for(int i=1;i<=n;i++) { scanf("%d",&a[i]); } int ans = 0; long long res = 0; for(int i=n;i>=1;) { res = res + a[i]; if(res >= 0) { ans++; i--; res = 0; } else { while(res < 0) { res = res + a[--i]; } ans++; i--; res = 0; } } printf("%d\n",ans); } return 0; }
相关文章推荐
- (HDU 5773)The All-purpose Zero <最长上升子序列 + 思维题> 多校训练4
- (HDU 5774)Where Amazing Happens <水题> 多校训练4
- HDU 5783 Divide the Sequence(水题)
- hdu 5783 Divide the Sequence【水题】
- HDU-5783 Divide the Sequence(贪心水题)
- HDU 5783 Divide the Sequence 思维题
- hdu 5783 Divide the Sequence(2016 Multi-University Training Contest 5——水题)
- HDOJ 5783 (2016多校联合训练 Training Contest 5) Divide the Sequence
- HDU 5783 Divide the Sequence 【贪心】
- hdu 5783 Divide the Sequence
- HDU 5783 Divide the Sequence
- 【HDU】5783 Divide the Sequence
- HDU_5783_DivideTheSequence(贪心)
- hdu 5783 Divide the Sequence 贪心
- (HDU 5922)Minimum’s Revenge 思维水题 <2016CCPC东北地区大学生程序设计竞赛 - 重现赛 >
- hdu-5783 Divide the Sequence(贪心)
- hdu 5783 Divide the Sequence
- hdu 5783——Divide the Sequence
- HDU 5783 Divide the Sequence
- (HDU 5924)Mr. Frog’s Problem 思维水题 <2016CCPC东北地区大学生程序设计竞赛 - 重现赛 >