题目大意
给你一串数组，按要求分割，使得每一段内的负数个数不超过2个，问最少分割组数以及每一组的个数
Sample
#1 IN
11
1 2 3 -4 -5 -6 5 -5 -6 -7 6
#1 OUT
3
5 3 3
#2 IN
5
0 -1 100 -1 0
#2 OUT
1
5
解题思路
最近感觉很不在状态，做水题看半天做必WA几次。虽然过了打算再整理一下
（烦~）
暴力啊！（我也是服了自己，这道题写了这么长的代码）
#include <bits/stdc++.h>
using namespace std;
const int maxn = 105;
int b[maxn];
int main()
{
std::ios::sync_with_stdio(false);
int n;
cin >> n;
vector<int> ve;
for (int i = 1; i <= n; i++)
{
int temp;
cin >> temp;
if (temp < 0)
{
ve.push_back(i);
}
}
int ans;
if (ve.size() % 2 != 0 || ve.size() == 0)
{
ans = ve.size() / 2 + 1;
}
else
{
ans = ve.size() / 2;
}
cout << ans << endl;
if (ans > 1)
{
int pre = 0, temp = 0, cont = 0, count = 0;
for (int i = 0; i < ve.size(); i++)
{
cont++;
if (cont > 1 && i == 1)
{
b[count++] = ve[i];
cont = 0;
}
else if (cont == 1 && i == ve.size() - 1)
{
b[count++] = n - ve[i-1];
cont = 0;
}
else if (cont > 1 && i == ve.size() - 1 && i < n - 1)
{
b[count++] = n - ve[i - 2];
cont = 0;
}
else if (cont > 1)
{
b[count++] = ve[i] - ve[i - 2];
cont = 0;
}
}
for (int i = 0; i < count; i++)
{
if (i == 0)
cout << b[i];
else
{
cout << " " << b[i];
}
}
cout << endl;
}
else{
cout << n << endl;
}
return 0;
}