C. Willem, Chtholly and Seniorious
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
— Willem…
— What’s the matter?
— It seems that there’s something wrong with Seniorious…
— I’ll have a look…
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.
Examples
input
10 10 7 9
output
2
1
0
3
input
10 10 9 9
output
1
1
3
3
题意：给你一个数组，所有处理都是随机的，对于4种操作做对应处理。
做法：emmmm，学到了一个新东西？如果把相同的数作为一个区间，有两种操作：1：将l,r设为一个数，2：把l,r的数都增加c，对于第二种操作就相当于多了两个区间，这样每次遍历所有的区间，复杂度在nlogn左右，，，然后我不知道怎么证明（题解也是这么说QAQ）；
#include<bits/stdc++.h>
#define ll long long
#define pii pair<int,ll>
using namespace std;
const int N= 1e5+100;
const int mod= 1e9+7;
int n,m,seed,mn;
int num[N];
map<int,pii> mp;
int rnd(){
int ret = seed;
seed = (7LL*seed+13)%mod;
return ret;
}
ll qpow(int x,int b,int p){
ll sum = 1;
ll now = x;
while(b){
if(b&1) sum = sum*now%p;
now = now*now%p;
b >>= 1;
}
return sum;
}
void solve(int op,int l,int r,int x,int y){
if(op == 1){
int now = 1;
while(now <= r){
pair<int,ll> &tmp = mp[now];
if(tmp.first >= l){
if(now < l){
mp[l] = tmp;
tmp.first = l-1;
}
else{
if(tmp.first <= r){
tmp.second += x;
}
else{
mp[r+1] = tmp;
tmp.first = r;
tmp.second += x;
}
}
}
now = tmp.first+1;
}
}
else if(op == 2){
int now = 1;
while(now <= r){
pair<int,ll> &tmp = mp[now];
if(tmp.first >= l){
if(now < l){
mp[l] = tmp;
tmp.first = l-1;
}
while(mp[now].first <= r){
now = mp[now].first+1;
}
mp[r+1] = mp[now];
mp[l] = make_pair(r,x);
now = l;
}
now = mp[now].first+1;
}
}
else if(op == 3){
int now = 1;
vector<pair<ll,int> > v;
while(now <= r){
pair<int,ll> &tmp = mp[now];
if(tmp.first >= l){
v.push_back({tmp.second,min(r,tmp.first)-max(l,now)+1});
}
now = tmp.first+1;
}
sort(v.begin(),v.end());
int bef = 0;
for(int i = 0;i < v.size();i ++){
bef += v[i].second;
if(bef >= x){
printf("%lld\n",v[i].first);
break;
}
}
}
else if(op == 4){
ll ans = 0;
int now = 1;
while(now <= r){
pair<int,ll> &tmp = mp[now];
if(tmp.first >= l){
ans =(ans+ 1LL*(min(r,tmp.first)-max(l,now)+1)*qpow((int)(tmp.second%y),x,y))%y;
}
now = tmp.first+1;
}
printf("%lld\n",ans);
}
}
int main(){
scanf("%d %d %d %d",&n,&m,&seed,&mn);
for(int i = 1;i <= n;i ++){
num[i] = rnd()%mn+1;
//cout <<i << ' '<< num[i] << endl;
mp[i] = make_pair(i,num[i]);
}
mp[n+1] = make_pair(n+1,0);
for(int i = 1;i <= m;i ++){
int op,l,r,x,y=0;
op = rnd()%4+1;
l = rnd()%n+1;
r = rnd()%n+1;
if(l > r) swap(l,r);
if(op == 3) x = rnd()%(r-l+1)+1;
else x = rnd()%mn+1;
if(op == 4) y = rnd()%mn+1;
//cout<<"!!" <<op << ' '<< l << ' '<< r << ' '<< x << ' '<< y << endl;
solve(op,l,r,x,y);
}
return 0;
}