DAY1小题

F

求逆序对的板子题

#include<cstdio>
#define ll long long
using namespace std;

const int maxn=1e5+5;
ll a[maxn],r[maxn],n;
ll ans=0;

void msort(ll s,ll t)
{
    if(s==t)
        return;
    ll mid=s+t>>1;
    msort(s,mid),msort(mid+1,t);
    ll i=s,j=mid+1,k=s;
    while(i<=mid&&j<=t)
        if(a[i]<=a[j])
            r[k++]=a[i++];
        else
            r[k++]=a[j++],ans+=(ll)mid-i+1;
    while(i<=mid)
        r[k]=a[i],k++,i++;
    while(j<=t)
        r[k]=a[j],k++,j++;
    for(int i=s; i<=t; i++)
        a[i]=r[i];
}
inline ll read()
{
    char ch=getchar();
    ll sum=0,k=1;
    while(ch<'0'||ch>'9')
    {
        if(ch=='-')
            k=-1;
        ch=getchar();
    }
    while(ch>='0'&&ch<='9')
        sum=sum*10+(ch^48),ch=getchar();
    return sum*k;
}
int main()
{
    scanf("%lld",&n);
    for(int i=1; i<=n; i++)
        a[i]=read();
    msort(1,n);
    printf("%lld
",ans);
    return 0;
}

E

unsigned long long等数据类型的定义方法的巧用

（直接long long居然也可以水到90分.）

#include<cstdio>
using namespace std;
int main()
{
    unsigned long long n;
    scanf("%llu",&n);
    printf("%llu",n*n);
    return 0;
}

D

单调队列

对于每一个数记录他是第几个放入的

放入时遇到比他小的元素就弹掉

这样保证了队首就是最大值

进行弹出操作时只要关注弹的次数是否等于队首元素id就可以了

（copy大佬的题解）

#include<cstdio>
#include<utility>
#include<deque>
#include<algorithm>
using namespace std;
int n,num,cnt;
deque<pair<int,int> >q;
void solu1(int x,int n)
{
    while(q.size() != 0 && q.back().first< x)
    {
        q.pop_back();
    }
    q.push_back(make_pair(x,n));
}
void solu2(int x)
{
    while(q.front().second <= x && q.size() != 0)
        q.pop_front();
}
int main()
{
    scanf("%d",&n);
    int opt;
    for(int i = 1; i <= n; i++)
    {
        scanf("%d",&opt);
        if(opt == 1)
        {
            int x;
            scanf("%d",&x);
            solu1(x,++num);
        }
        if(opt == 2)
        {
            solu2(++cnt);
        }
        if(q.size() == 0)
            printf("-1
");
        else
            printf("%d
",q.front().first);
        }
    return 0;
}

C

(大概是我写两个cmp是不对的

（一个就够了qwq）

（于是爆0.

#include<cstdio>
#include<algorithm>
#include<deque>
using namespace std;

struct shu
{
    int x,y;
};
shu c[maxn+5];
int n;

bool cmp(shu a,shu b)
{
    if(a.x==b.x) return a.y<b.y;
    return a.x<b.x;
}

int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        scanf("%d%d",&c[i].x,&c[i].y);
    }
    sort(c+1,c+n+1,cmp);
    for(int i=1;i<=n;i++)
    {
        printf("%d %d
",c[i].x,c[i].y);
    }
    return 0;
}

B

双向队列的基本操作

#include<cstdio>
#include<algorithm>
#include<deque>
using namespace std;
int n,opt;
deque<int>q;
int main()
{
    scanf("%d",&n);
    for(int i = 1; i <= n; i++)
    {
        scanf("%d",&opt);
        if(opt == 1)
        {
            int x;
            scanf("%d",&x);
            if(q.size() == 0)
                printf("-1
");
            else
                printf("%d
",q.back());
            q.push_back(x);
        }
        else if(opt == 2)
        {
            if(q.size() == 0)
                printf("-1
");
            else
            {
                printf("%d
",q.back());
                q.pop_back();
            }
        }
        else if(opt == 3)
        {
            int x,y;
            scanf("%d%d",&x,&y);
            if(q.size()<=x)
                printf("-1
");
            else
            {
                printf("%d
",q[x]);
                q[x]=y;
            }
        }
        else if(opt == 4)
        {
            int x;
            scanf("%d",&x);
            if(q.size() == 0)
                printf("-1
");
            else
                printf("%d
",q.front());
            q.push_front(x);
        }
        else if(opt == 5)
        {
            if(q.size() == 0)
                printf("-1
");
            else
            {
                printf("%d
",q.front());
                q.pop_front();
            }
        }
    }
    return 0;
}

A

开两个栈，一个用来模拟真实的栈，另一个求最大值。

每当加入新元素时

如果该元素大于等于栈顶元素 就压进去

弹出元素时，如果真实栈的弹出元素和最大值栈的栈顶元素相等，就弹出。

（qxt好gou啊.

//实现一个栈，支持push和pop，每次执行完毕后要输出栈内元素的最大值
#include<cstdio>
#include<stack>
using namespace std;
stack<int> q,pq;
int n;
void solu1(int x)
{
    q.push(x);
    if(pq.size() == 0 || pq.top() <= x)
        pq.push(x);
}
void solu2()
{
    if(q.size())
    {
        if(pq.size())
            if(q.top() == pq.top())
            {
                q.pop();
                pq.pop();
            }
            else
                q.pop();
        else
            q.pop();
    }
}
int main()
{
    scanf("%d",&n);
    int opt;
    for(int i = 1; i <= n; i++)
    {
        scanf("%d",&opt);
        if(opt == 1)
        {
            int x;
            scanf("%d",&x);
            solu1(x);
        }
        if(opt == 2)
        {
            solu2();
        }

        if(pq.size() == 0)
            printf("-1
");
        else
            printf("%d
",pq.top());
    }
    return 0;
}