Zhongshan Day10-- popularity

Today is stupid day, Gufen 230, take-110. Which T2,4 not here just say that Italy and brief thoughts.

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T1: Template title

 

This problem is relatively simple, that is, when read by a row of buckets, all the factors of the digital read one, then, again descending sweep, to look at whether the number i is a multiple of k, then the output It can be.

See Code:

#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<cstdio>
using namespace std;
int n,v,k=1,maxx=-0x3f3f3f,a[100001],t[100001],b[100001];
void g(int x)
{
    for(int i=1;i*i<=x;i++)
    {
        if(x%i==0)
        {
            t[i]++;
            t[x/i]++;
            if(i*i==x)
            t[i]--;
        }
    }
}
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&v);
        g(v);
        if(v>maxx)
        maxx=v;
    }
    for(int i=maxx;i>=1;i--)
    {
        while(t[i]>=k)
        {
            b[k]=i;
            k++;
        }
    }
    for(int i=1;i<=n;i++)
    printf("%d\n",b[i]);
    return 0;
}

Good question indeed! ! !

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T2: stretching

Because it is the n- ² , but I will only simulate violence, so too.

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T3: clues

Idea: when encounter left parenthesis, with a large root pile up savings. Then when the encounter left parenthesis, pressed into a small heap with, if left stack is empty, then pop the right piles top. Otherwise, left piles of top pop-up. Finally, if the pile is still not left empty, then left empty heap. This is clearly greedy correctness: When the left and right parenthesis encounter with a left parenthesis, priority take a big price to match. And if there is no left parenthesis, right parenthesis is not the minimum cost to use. Finally, how if left bracket, then all the pop-up.

See Code:

#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<queue>
#define ll long long
using namespace std;
priority_queue<int,vector<int>,less<int> >q1;
priority_queue<int,vector<int>,greater<int> >q2;
ll n,k,k1=1,red,ans;
char a[100001];
int main()
{
    freopen("assassin.in","r",stdin);
    freopen("assassin.out","w",stdout);
    scanf("%lld",&n);
    for(ll i=1;i<=n;i++)
    scanf(" %c",&a[i]);
    for(ll i=1;i<=n;i++)
    {
        scanf("%lld",&red);
        if(a[i]=='(')
        {
            q1.push(red);
        }
        else
        {
            q2.push(red);
            if(!q1.empty())
            {
                q1.pop();
                k++;
            }
            else
            {
                ans+=q2.top();
                q2.pop();
            }
        }
    }
    while(!q1.empty())
    {
        ans+=q1.top();
        q1.pop();
    }
    printf("%lld",ans);
    return 0;
}

好题哉！！！

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T4:自由

 

据大佬们说，此题是动归的斜率优化，当然数据有点小$n^{2}$也可以过。

但是……最基础的$n^{3}$我都不会。所以如清风般略过。

好题哉！！！