杭州电子科技大学2018多校测试第一场

杭州电子科技大学2018多校测试第一场

A Maximum Multiple

打表规律题

#include<bits/stdc++.h>
using namespace std;
/*int work(int k)
{
    int ans=0,pi=0,pj=0,pt=0;
    for (int i=1; i<k; i++) if (k%i==0)
    {
        for (int j=1; j<(k-i); j++) if (k%j==0)
        {
            int t=k-i-j;
            if (k%t==0)
            {
                if (i*j*t>ans)
                {
                    ans=i*j*t;
                    pi=i; pj=j; pt=t;
                }
            }
        }
    }
    cout<<k<<" "<<pi<<" "<<pj<<" "<<pt<<" "<<ans<<endl;
}
int main()
{
    for (int i=1; i<=100; i++)
    {
        work(i);
    }
}*/
int main()
{
    int T;
    scanf("%d",&T);
    while (T--)
    {
        long long  x,ans;
        scanf("%lld",&x);
        if (x%3==0)
        {
            ans=(x/3)*(x/3)*(x/3);
        }
        else if (x%4==0)
        {
            ans=(x/2)*(x/4)*(x/4);
        }
        else ans=-1;
        printf("%lld
",ans);
    }
}

B Balanced Sequence

贪心

/*
    第一反应感觉可以模拟贪心一下
    处理完当前段必加答案之后感觉是dp
    之后发现DP就会很麻烦
    看题解发现贪心相关
*/

#include<bits/stdc++.h>
#define N 1000005
using namespace std;
char ch[N];
struct Node{
    int left,right;
}a[N];
int n;
/*bool cmp(Node a,Node b)
{
    if (a.left>=a.right && b.left<=b.right) return true;
    else if (a.left>=a.right && b.left>=b.right) return a.right<b.right;
    else if (a.left<=b.right && b.left>=b.right) return false;
    else if (a.left<=b.right && b.left<=b.right) return a.left>b.left;
    return true;
}*/
bool cmp(Node a, Node b){
    if(a.left > a.right)    //判断'('大于')'时，如果a，b都满足，则谁的')'小谁排前面，否则a排前面
        return b.left > b.right? a.right<b.right:true;
    return b.left <= b.right?a.left>b.left:false;
}

int main()
{
    int T;
    scanf("%d",&T);
    while (T--)
    {
        scanf("%d",&n);
        int preleft=0,preright=0,ans=0;
        for (int i=1; i<=n; i++)
        {
            scanf("%s",ch);
            int len=strlen(ch);
            int nowleft=0,nowright=0,nowans=0;
            for (int i=0; i<len; i++)
            {
                if (ch[i]=='(') nowleft++;
                else
                {
                    if (nowleft>0) nowleft--,nowans++;
                    else nowright++;
                }
            }
            ans+=nowans;
            a[i].left=nowleft; a[i].right=nowright;
        }
        sort(a+1,a+n+1,cmp);
        preleft=a[1].left; preright=a[1].right;
        for (int i=2; i<=n; i++)
        {
            int tmp=min(preleft,a[i].right);
            ans+=tmp;
            preleft=preleft+a[i].left-tmp; preright=preright+a[i].right-tmp;
        }
        printf("%d
",ans*2);
    }

    return 0;
}

C Triangle Partition

排序

#include<bits/stdc++.h>
#define N 30005
using namespace std;
struct Node{
    int x,y,pos;
}a[N];
int n;
bool cmp(Node A,Node B)
{
    if (A.x!=B.x) return A.x<B.x;
    else return A.y<B.y;
}
int main()
{
    int T;
    scanf("%d",&T);
    while (T--)
    {
        scanf("%d",&n);
        n=n*3;
        for (int i=1; i<=n; i++)
        {
            scanf("%d%d",&a[i].x,&a[i].y);
            a[i].pos=i;
        }
        sort(a+1,a+n+1,cmp);
        for (int i=1; i<n; i++)  printf("%d ",a[i].pos);
        printf("%d
",a[n].pos);
    }
    return 0;
}

D Distinct Values

去重+贪心+模拟

#include<bits/stdc++.h>
#define N 1000005
using namespace std;
struct Node {
    int l,r;
}a[N],b[N];
int ans[N];
bool vis[N];
int n,m;
bool cmp(Node a, Node b)
{
    if (a.l==b.l) return a.r>b.r; else return a.l<b.l;
}
int main()
{
    int T;
    scanf("%d",&T);
    while (T--)
    {
        scanf("%d%d",&n,&m);
        for (int i=1; i<=m; i++)
        {
            scanf("%d%d",&a[i].l,&a[i].r);
        }
        sort(a+1,a+m+1,cmp);
        int maxn=0,tot=0;
        for (int i=1; i<=m; i++)
        {
            if (maxn>=a[i].r) continue;
            else
            {
                b[++tot]=a[i];
               // cout<<"           ssssss "<<a[i].l<<" "<<a[i].r<<endl;
                maxn=a[i].r;
            }
        }
        //cout<<tot<<endl;
        for (int i=1; i<=n; i++) ans[i]=1;
        int pre=0;
        for (int i=1; i<=tot; i++)
        {
            if (b[i].l>pre)
            {
                int tmp=0;
                for (int j=b[i].l; j<=b[i].r; j++) ans[j]=++tmp;
                pre=b[i].r;
            }
            else
            {
                for (int j=b[i].l; j<=pre; j++)
                {
                    vis[ans[j]]=1;
                }
                int tmp=1;
                for (int j=pre+1; j<=b[i].r; j++)
                {
                    while (vis[tmp]!=0) tmp++;
                    ans[j]=tmp;
                    vis[tmp]=1;
                }
                for (int j=b[i].l; j<=b[i].r; j++)
                {
                    vis[ans[j]]=0;
                }
                pre=b[i].r;
            }
        }
        for (int i=1; i<n; i++) printf("%d ",ans[i]);
        printf("%d
",ans[n]);
    }
    return 0;
}