链接：https://codeforces.com/contest/2025/problem/D
题目：
思路：
动态规划。dp[i]记录当前0分配了i个给智力所能达到的最大分数。利用strength[N],intel[N]表示小于等于i的个数，所以加上前缀和赋值给dp[i],然后清空两个数组，方便这个零到下个零的这段。
代码：
#include<iostream>
using namespace std;
#define int long long
#define IOS ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
const int N = 2e6 + 10;
int strength[N], intel[N];
int nums[N],dp[N];
int n, m;
signed main()
{
IOS;
cin >> n >> m;
for (int i = 1; i <= n; i++)
{
cin >> nums[i];
}
int dfn = 0;
for (int i = 1; i <= n; i++)
{
if (nums[i] > 0)
{
intel[nums[i]]++;
}
else if (nums[i] < 0)
{
strength[-nums[i]]++;
}
if(nums[i] == 0 or i == n)
{
for (int j = 1; j <= dfn; j++)
{
strength[j] += strength[j - 1];
intel[j] += intel[j - 1];
}
for (int j = 0; j <= dfn; j++)
{
dp[j] += intel[j] + strength[dfn - j];
}
dfn++;
for (int j = dfn; j > 0; j--)
{
dp[j] = max(dp[j], dp[j - 1]);
}
for (int j = 1; j <= dfn; j++)
{
intel[j] = strength[j] = 0;
}
}
}
int ans = 0;
for (int i = 1; i <= dfn; i++)ans = max(dp[i], ans);
cout << ans;
return 0;
}