标签：处理 copy bsp bytes force second tco standard over
题目链接: http://codeforces.com/contest/1343/problem/C
C. Alternating Subsequence
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
Recall that the sequence b is a a subsequence of the sequence a if b can be derived from a by removing zero or more elements without changing the order of the remaining elements. For example, if a=[1,2,1,3,1,2,1], then possible subsequences are: [1,1,1,1], [3] and [1,2,1,3,1,2,1], but not [3,2,3] and [1,1,1,1,2].
You are given a sequence a consisting of n positive and negative elements (there is no zeros in the sequence).
Your task is to choose maximum by size (length) alternating subsequence of the given sequence (i.e. the sign of each next element is the opposite from the sign of the current element, like positive-negative-positive and so on or negative-positive-negative and so on). Among all such subsequences, you have to choose one which has the maximum sum of elements.
In other words, if the maximum length of alternating subsequence is k then your task is to find the maximum sum of elements of some alternating subsequence of length k.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1≤t≤104) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1≤n≤2⋅105) — the number of elements in a. The second line of the test case contains n integers a1,a2,…,an (−109≤ai≤109,ai≠0), where ai is the i-th element of a.
It is guaranteed that the sum of n over all test cases does not exceed 2⋅105 (∑n≤2⋅105).
Output
For each test case, print the answer — the maximum sum of the maximum by size (length) alternating subsequence of a.
Example
inputCopy
4
5
1 2 3 -1 -2
4
-1 -2 -1 -3
10
-2 8 3 8 -4 -15 5 -2 -3 1
6
1 -1000000000 1 -1000000000 1 -1000000000
outputCopy
2
-1
6
-2999999997
题意：就是给你一串数你需要找出其中的子字符串，并且这个子字符串是正负交替的，并且尽量让这个子字符串的和最大。
思路：这样的处理，就栈用着最方便了，对前面的数是正是负稍加判断，分别处理就好了。
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int main()
{
int t;
cin>>t;
while(t--)
{
stack<int> s;
int n;
cin>>n;
int flag=1;
int k;
cin>>k;
s.push(k); // 先插入第一个
if(k<0) // 先判断下第一个
{
flag=0; // 当前位为负数
}
else
{
flag=1; // 当前位为正数
}
for(int i=1;i<n;i++)
{
cin>>k;
if(flag==0) // 当前位为负数
{
if(k<0&&k>s.top()) // 后一位为负的并且比当前栈顶大，替换掉
{
s.pop();
s.push(k);
}
if(k>0) // 与其符号相反，压入即可
{
s.push(k);
flag=1;
}
}
else // 当前位为正数
{
if(k>0&&k>s.top()) // 后一位为正的并且比当前栈顶大，替换掉
{
s.pop();
s.push(k);
}
if(k<0) // 与其符号相反，压入即可
{
s.push(k);
flag=0;
}
}
}
ll num=0;
while(!s.empty()) // 累加求和，输出即可
{
//cout<<s.top()<<" ";
num+=s.top();
s.pop();
}
cout<<num<<endl; //输出即可
}
return 0;
}
共同加油！
坚持！
Alternating Subsequence _牛哄哄的柯南
标签：处理 copy bsp bytes force second tco standard over
原文地址：https://www.cnblogs.com/Keafmd/p/12933945.html
