Codeforces Round #449 (Div. 1)C - Willem, Chtholly and Seniorious

ODT(主要特征就是推平一段区间)

其实就是用set来维护三元组,因为数据随机所以可以证明复杂度不超过O(NlogN),其他的都是暴力维护

主要操作是split,把区间分成两个,用lowerbound,

有两点需要注意1.set里的东西不能改,所以变成了mutable(可改的const),2.s.insert返回pair类型,first是插入点的迭代器

= =sort里的比较函数ll没开wa了好久

精简版:

//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
//#pragma GCC optimize("unroll-loops")
#include<bits/stdc++.h>
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define pi acos(-1.0)
#define ll long long
#define vi vector<int>
#define mod 1000000007
#define ld long double
#define C 0.5772156649
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define pll pair<ll,ll>
#define pil pair<int,ll>
#define pli pair<ll,int>
#define pii pair<int,int>
#define cd complex<double>
#define ull unsigned long long
#define base 1000000000000000000
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define fio ios::sync_with_stdio(false);cin.tie(0)
inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}
inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
//inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}
inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=ans*a%c;a=a*a%c,b>>=1;}return ans;}

using namespace std;

const double eps=1e-8;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int N=200000+10,maxn=50000+10,inf=0x3f3f3f3f;

struct ODT{
    struct node{
        int l,r;mutable ll val;
        node(int _l,int _r = -1,ll _val = 0):l(_l),r(_r),val(_val){}
        bool operator <(const node&rhs)const{
            return l<rhs.l;
        }
    };
    set<node>s;
    typedef set<node>::iterator sit;
    void ins(int l,int r,ll val){s.insert(node(l,r,val));}
    sit split(int pos)
    {
        sit p=s.lower_bound(node(pos));
        if(p != s.end() && pos == p->l)return p;
        p--;node te=*p;
        s.erase(p);
        s.insert(node(te.l,pos-1,te.val));
        return s.insert(node(pos,te.r,te.val)).fi;
    }
    void add(int l,int r,ll v)
    {
        sit x=split(l),y=split(r+1);
        for(;x!=y;x++)x->val += v;
    }
    void color(int l,int r,ll val)
    {
        sit x= split(l),y=split(r+1);
        s.erase(x,y);
        s.insert(node(l,r,val));
    }
    void kth(int l,int r,ll k)
    {
        sit x = split(l),y = split(r+1);
        vector<pli>v;
        for(;x!=y;x++)
            v.pb(mp(x->val,x->r - x->l + 1));
        sort(v.begin(),v.end(),[](pli a,pli b){return a.fi>b.fi;});
        for(int i=(int)v.size()-1;i>=0;i--)
        {
            if(k<=v[i].se)
            {
                printf("%lld\n",v[i].fi);
                return ;
            }
            k-=v[i].se;
        }
    }
    void sum(int l,int r,ll a,ll b)
    {
        sit x = split(l),y=split(r+1);
        ll ans=0;
        for(;x!=y;x++)
        {
            ans+=1ll*qp(x->val % b,a,1ll*b)*(x->r - x->l + 1)%b;
            if(ans>=b)ans-=b;
        }
        printf("%lld\n",ans);
    }
    void debug()
    {
        for(auto x:s)
            printf("%d %d %d\n",x.l,x.r,x.val);
        puts("");
    }
}o;
ll n,m,seed,vmax,a[N];
ll rnd()
{
    ll ret = seed;
    seed = (seed * 7ll + 13)%1000000007;
    return ret;
}
int main()
{
//    o.ins(1,1,3);o.ins(2,2,4),o.ins(3,5,6);
//    o.add(2,3,2);
//    o.debug();
    scanf("%lld%lld%lld%lld",&n,&m,&seed,&vmax);
    for(int i=1;i<=n;i++)
    {
        a[i] = (rnd()%vmax) + 1;
        o.ins(i,i,a[i]);
    }
    o.ins(n+1,n+1,0);
    int res=0;
    for(int i=1;i<=m;i++)
    {
        int op = (rnd()%4) + 1;
        int l = (rnd()%n) + 1;
        int r = (rnd()%n) + 1;
        ll x,y;
        if (l > r)swap(l, r);
        if (op == 3)x = (rnd()%(r - l + 1)) + 1;
        else x = (rnd()%vmax) + 1;
        if (op == 4)y = (rnd()%vmax) + 1;
        if(op==1)o.add(l,r,x);
        else if(op==2)o.color(l,r,x);
        else if(op==3)o.kth(l,r,x);
        else o.sum(l,r,x,y);
//        o.debug();
//        printf("%d %d %d %d %d\n",op,l,r,x,y);
    }
    return 0;
}
/********************

********************/

自已刚开始意淫的一个写法:

//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
//#pragma GCC optimize("unroll-loops")
#include<bits/stdc++.h>
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define pi acos(-1.0)
#define ll long long
#define vi vector<int>
#define mod 1000000007
#define ld long double
#define C 0.5772156649
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define pll pair<ll,ll>
#define pil pair<int,ll>
#define pli pair<ll,int>
#define pii pair<int,int>
#define cd complex<double>
#define ull unsigned long long
#define base 1000000000000000000
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define fio ios::sync_with_stdio(false);cin.tie(0)
inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}
inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}
inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=ans*a%c;a=a*a%c,b>>=1;}return ans;}

using namespace std;

const double eps=1e-8;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int N=200000+10,maxn=50000+10,inf=0x3f3f3f3f;

struct ODT{
    struct node{
        int l,r;mutable ll val;
        bool operator <(const node&rhs)const{
            return l<rhs.l || l==rhs.l&&r<rhs.r;
        }
    };
    set<node>s;
    void split(int pos)
    {
        if(pos<=0)return ;
        auto p=s.upper_bound(node{pos,inf,INF});
        if(p==s.begin())return ;
        p--;
//        printf("--------%d %d %d\n",p->l,p->r,p->val);
        if(pos < p->l || pos >= p->r)return ;
        node te=*p;
        s.erase(te);
        s.insert({te.l,pos,te.val});
        s.insert({pos+1,te.r,te.val});
    }
    void add(int l,int r,ll v)
    {
        split(l-1),split(r);
        auto x=s.lower_bound(node{l,-inf,-INF});
        auto y=s.lower_bound(node{r,inf,INF});
        for(;x!=y;x++)x->val += v;
    }
    void color(int l,int r,ll val)
    {
        split(l-1);split(r);
        auto x=s.lower_bound(node{l,-inf,-INF});
        auto y=s.lower_bound(node{r,inf,INF});
//        printf("%d %d %d %d\n",x->l,x->r,y->l,y->r);
        s.erase(x,y);
        s.insert({l,r,val});
    }
    void kth(int l,int r,ll k)
    {
        split(l-1);split(r);
        auto x=s.lower_bound(node{l,-inf,-INF});
        auto y=s.lower_bound(node{r,inf,INF});
        vector<pli>v;
        for(;x!=y;x++)
            v.pb(mp(x->val,x->r - x->l + 1));
        sort(v.begin(),v.end(),[](pli a,pli b){return a.fi>b.fi;});
        for(int i=(int)v.size()-1;i>=0;i--)
        {
            if(k<=v[i].se)
            {
                printf("%lld\n",v[i].fi);
                return ;
            }
            k-=v[i].se;
        }
    }
    void sum(int l,int r,ll a,ll b)
    {
        split(l-1);split(r);
        auto x=s.lower_bound(node{l,-inf,-INF});
        auto y=s.lower_bound(node{r,inf,INF});
        ll ans=0;
        for(;x!=y;x++)
        {
            ans+=1ll*qp(x->val % b,a,1ll*b)*(x->r - x->l + 1)%b;
            if(ans>=b)ans-=b;
        }
        printf("%lld\n",ans);
    }
    void debug()
    {
        for(auto x:s)
            printf("%d %d %d\n",x.l,x.r,x.val);
        puts("");
    }
}o;
ll n,m,seed,vmax,a[N];
ll rnd()
{
    ll ret = seed;
    seed = (seed * 7ll + 13)%1000000007;
    return ret;
}
int main()
{
    scanf("%lld%lld%lld%lld",&n,&m,&seed,&vmax);
    for(int i=1;i<=n;i++)
    {
        a[i] = (rnd()%vmax) + 1;
        o.s.insert({i,i,a[i]});
    }
//    o.debug();
    for(int i=1;i<=m;i++)
    {
        int op = (rnd()%4) + 1;
        int l = (rnd()%n) + 1;
        int r = (rnd()%n) + 1;
        ll x,y;
        if (l > r)swap(l, r);
        if (op == 3)x = (rnd()%(r - l + 1)) + 1;
        else x = (rnd()%vmax) + 1;
        if (op == 4)y = (rnd()%vmax) + 1;
        if(op==1)o.add(l,r,x);
        else if(op==2)o.color(l,r,x);
        else if(op==3)o.kth(l,r,x);
        else o.sum(l,r,x,y);
//        o.debug();
//        printf("%d %d %d %d %d\n",op,l,r,x,y);
    }
    return 0;
}
/********************

********************/