CodeForces - 896C Willem, Chtholly and Seniorious【ODT珂朵莉树】

版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/niiick/article/details/83062548

Time limit 2000 ms
Memory limit 262144 kB

Seniorious is made by linking special talismans in particular order.

After over 500 years, the carillon is now in bad condition, so Willem decides to examine it thoroughly.

Seniorious has n pieces of talisman. Willem puts them in a line, the i-th of which is an integer ai.

In order to maintain it, Willem needs to perform m operations.

There are four types of operations:

1 l r x: For each i such that l ≤ i ≤ r, assign ai + x to ai.
2 l r x: For each i such that l ≤ i ≤ r, assign x to ai.
3 l r x: Print the x-th smallest number in the index range [l, r], i.e. the element at the x-th position if all the elements ai such that l ≤ i ≤ r are taken and sorted into an array of non-decreasing integers. It’s guaranteed that 1 ≤ x ≤ r - l + 1.
4 l r x y: Print the sum of the x-th power of ai such that l ≤ i ≤ r, modulo y, i.e. .

Input

The only line contains four integers n, m, seed, vmax (1 ≤ n, m ≤ 105, 0 ≤ seed < 109 + 7, 1 ≤ vmax ≤ 109).

The initial values and operations are generated using following pseudo code:

def rnd():

    ret = seed
    seed = (seed * 7 + 13) mod 1000000007
    return ret

for i = 1 to n:

    a[i] = (rnd() mod vmax) + 1

for i = 1 to m:

    op = (rnd() mod 4) + 1
    l = (rnd() mod n) + 1
    r = (rnd() mod n) + 1

    if (l > r): 
         swap(l, r)

    if (op == 3):
        x = (rnd() mod (r - l + 1)) + 1
    else:
        x = (rnd() mod vmax) + 1

    if (op == 4):
        y = (rnd() mod vmax) + 1

Here op is the type of the operation mentioned in the legend.

Output

For each operation of types 3 or 4, output a line containing the answer.

---

题目分析

毒瘤ODT的源头啊

骗分导论–ODT珂朵莉树

区间Kth直接暴力取出对应区间排序就好

#define pir pair<int,int>//前一个记录值，后一个记录出现次数
int kth(int ll,int rr,int k)
{
    vector<pir> vec;
    IT itr=split(rr+1),itl=split(ll);
    
    for(;itl!=itr;++itl)
    vec.push_back( pir(itl->val,itl->rr-itl->ll+1) );
    sort(vec.begin(),vec.end());//全部存下来排序就好
    
    for(vector<pir>::iterator it=vec.begin();it!=vec.end();++it)
    {
        k-=it->second;
        if(k<=0) return it->first;
    }
    return -1;
}

另外注意爆long long的精度问题

---

#include<iostream>
#include<cmath>
#include<algorithm>
#include<vector>
#include<cstring>
#include<cstdio>
#include<set>
using namespace std;
typedef long long lt;
#define IT set<node>::iterator
#define pir pair<lt,int>
#define mkp(x,y) make_pair(x,y)

lt read()
{
    lt f=1,x=0;
    char ss=getchar();
    while(ss<'0'||ss>'9'){if(ss=='-')f=-1;ss=getchar();}
    while(ss>='0'&&ss<='9'){x=x*10+ss-'0';ss=getchar();}
    return f*x;
}

lt seed,vmax;
lt rnd()
{
    lt res=seed;
    seed=(seed*7+13)%1000000007;
    return res;
}

const int maxn=100010;
int n,m;
lt a[maxn];
struct node
{
    int ll,rr;
    mutable lt val;
    node(int L,int R=-1,lt V=0): ll(L), rr(R), val(V) {}
    bool operator < (const node& tt)const { return ll<tt.ll;}
};
set<node> st;

lt qpow(lt a,lt k,lt p)
{
    lt res=1; a%=p;
    while(k>0){
        if(k&1) res=(res*a)%p;
        a=(a*a)%p; k>>=1;
    }
    return res;
}

IT split(int pos)
{
    IT it=st.lower_bound(node(pos));
    if(it!=st.end()&&it->ll==pos) return it;
    --it;
    int ll=it->ll,rr=it->rr;
    lt val=it->val;
    st.erase(it);
    st.insert(node(ll,pos-1,val));
    return st.insert(node(pos,rr,val)).first;
}

void assign(int ll,int rr,lt val)
{
    IT itr=split(rr+1),itl=split(ll);
    st.erase(itl,itr);
    st.insert(node(ll,rr,val));
}

void add(int ll,int rr,lt val)
{
    IT itr=split(rr+1),itl=split(ll);
    for(;itl!=itr;++itl) itl->val+=val;
}

lt kth(int ll,int rr,int k)
{
    vector<pir> vec;
    IT itr=split(rr+1),itl=split(ll);
    for(;itl!=itr;++itl)
    vec.push_back(pir(itl->val,itl->rr-itl->ll+1));
    sort(vec.begin(),vec.end());
    for(vector<pir>::iterator it=vec.begin();it!=vec.end();++it)
    {
        k-=it->second;
        if(k<=0) return it->first;
    }
    return -1;
}

lt qsum(int ll,int rr,lt x,lt y)
{
    lt res=0;
    IT itr=split(rr+1),itl=split(ll);
    for(;itl!=itr;++itl)
    res+=(qpow(itl->val,x,y)*((itl->rr-itl->ll+1)%y))%y,res%=y;
    return res;
}

int main()
{
    n=read();m=read();
    seed=read();vmax=read();
    
    for(int i=1;i<=n;++i)
    {
        a[i]=(rnd()%vmax)+1;
        st.insert(node(i,i,a[i]));
    }
    
    for(int i=1;i<=m;++i)
    {
        int op=(rnd()%4)+1;
        int ll=(rnd()%n)+1,rr=(rnd()%n)+1;
        lt x,y;

    	if(ll>rr) swap(ll,rr);
    	if(op==3) x=(rnd()%(rr-ll+1))+1;
    	else x=(rnd()%vmax)+1;
    	if(op==4) y=(rnd()%vmax)+1;
    	
    	if(op==1) add(ll,rr,x);
    	else if(op==2) assign(ll,rr,x);
    	else if(op==3) printf("%lld\n",kth(ll,rr,x));
    	else if(op==4) printf("%lld\n",qsum(ll,rr,x,y));
    }
    return 0;
}