标签：Willem Chtholly itl int ll Seniorious rnd seed mod
Willem, Chtholly and Seniorious
https://codeforces.com/contest/897/problem/E
time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard 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.
.
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.
OutputFor each operation of types 3 or 4, output a line containing the answer.
Examples input Copy10 10 7 9output Copy
2input Copy
1
0
3
10 10 9 9output Copy
1Note
1
3
3
In the first example, the initial array is {8, 9, 7, 2, 3, 1, 5, 6, 4, 8}.
The operations are:
- 2 6 7 9
- 1 3 10 8
- 4 4 6 2 4
- 1 4 5 8
- 2 1 7 1
- 4 7 9 4 4
- 1 2 7 9
- 4 5 8 1 1
- 2 5 7 5
- 4 3 10 8 5
ODT模板题
参考博客：https://blog.csdn.net/niiick/article/details/83062256
1 #include<bits/stdc++.h>
2 using namespace std;
3 #define lson l,mid,rt<<1
4 #define rson mid+1,r,rt<<1|1
5 #define IT set<node>::iterator
6 #define sqr(x) ((x)*(x))
7 #define pb push_back
8 #define eb emplace_back
9 #define maxn 100005
10 #define eps 1e-8
11 #define pi acos(-1.0)
12 #define rep(k,i,j) for(int k=i;k<j;k++)
13 typedef long long ll;
14 typedef pair<int,int> pii;
15 typedef pair<ll,ll>pll;
16 typedef pair<ll,int> pli;
17 typedef pair<pair<int,string>,pii> ppp;
18 typedef unsigned long long ull;
19 const long long MOD=1e9+7;
20 const double oula=0.57721566490153286060651209;
21 using namespace std;
22
23 struct node{
24 int l,r;
25 mutable ll val;
26 node(int L,int R=-1,ll V=0):l(L),r(R),val(V){}
27 bool operator<(const node& t)const {
28 return l<t.l;
29 }
30 };
31 set<node>se;
32
33 ll ksm(ll a,ll b,ll mod){
34 ll ans=1;
35 a%=mod;
36 while(b){
37 if(b&1){
38 ans=(ans*a)%mod;
39 }
40 b>>=1;
41 a=(a*a)%mod;
42 }
43 return ans;
44 }
45
46 IT split(int pos){
47 IT it=se.lower_bound(node(pos));
48 if(it!=se.end()&&it->l==pos) return it;
49 it--;
50 int L=it->l,R=it->r;
51 ll V=it->val;
52 se.erase(it);
53 se.insert(node(L,pos-1,V));
54 return se.insert(node(pos,R,V)).first;///返回L==pos的迭代器
55 }
56
57 void assign(int l,int r,ll v){
58 IT itr=split(r+1),itl=split(l);
59 se.erase(itl,itr);///左闭右开
60 se.insert(node(l,r,v));
61 }
62
63 void add(int l,int r,ll v){
64 IT itr=split(r+1);
65 for(IT itl=split(l);itl!=itr;itl++){
66 itl->val+=v;
67 }
68 }
69
70 ll Rank(int l,int r,int v){
71 IT itr=split(r+1);
72 vector<pli>tmp;
73 for(IT itl=split(l);itl!=itr;itl++){
74 tmp.pb({itl->val,itl->r-itl->l+1});
75 }
76 sort(tmp.begin(),tmp.end());
77 for(int i=0;i<tmp.size();i++){
78 v-=tmp[i].second;
79 if(v<=0){
80 return tmp[i].first;
81 }
82 }
83 }
84
85 ll query(int l,int r,ll x,ll y){
86 IT itr=split(r+1);
87 ll ans=0;
88 for(IT itl=split(l);itl!=itr;itl++){
89 ans=(ans+((ksm(itl->val,x,y)*(itl->r-itl->l+1))%y))%y;
90 }
91 return ans;
92 }
93
94 ll n,m,seed,vmax;
95
96 ll rnd(){
97 ll ans=seed;
98 seed=(seed*7+13)%MOD;
99 return ans;
100 }
101
102
103 int main(){
104 std::ios::sync_with_stdio(false);
105
106 cin>>n>>m>>seed>>vmax;
107 ll x,y;
108 for(int i=1;i<=n;i++){
109 x=(rnd()%vmax)+1;
110 se.insert(node(i,i,x));
111 }
112 int opt,L,R;
113 for(int i=1;i<=m;i++){
114 opt=(rnd()%4)+1;
115 L=(rnd()%n)+1;
116 R=(rnd()%n)+1;
117 if(L>R) swap(L,R);
118 if(opt==3){
119 x=(rnd()%(R-L+1))+1;
120 }
121 else{
122 x=(rnd()%vmax)+1;
123 }
124 if(opt==4){
125 y=(rnd()%vmax)+1;
126 }
127 if(opt==1){
128 add(L,R,x);
129 }
130 else if(opt==2){
131 assign(L,R,x);
132 }
133 else if(opt==3){
134 cout<<Rank(L,R,x)<<endl;
135 }
136 else{
137 cout<<query(L,R,x,y)<<endl;
138 }
139 }
140 }
View Code
标签：Willem,Chtholly,itl,int,ll,Seniorious,rnd,seed,mod 来源： https://www.cnblogs.com/Fighting-sh/p/10584244.html
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