Prelude
ODT这个东西真是太好用了，以后写暴力骗分可以用，写在这里mark一下。
题目链接：ヽ(✿ﾟ▽ﾟ)ノ
Solution
先把原题解贴在这里：(ﾉ*･ω･)ﾉ
简单地说，因为数据是全部随机的，所以一定会有特别多的区间set，就会有很多数字相同，那么我们暴力把相同的数字合并成一个点，合并完之后数组就变得很短，然后对于询问暴力做就可以了。
具体复杂度是什么我也不会证明qwq，所以就把由乃的题解贴上来了qwq，感觉很靠谱的样子。
题解中给的是用STL的set维护缩点后的数组，我感觉不是很好写，就直接用数组维护了。
ddd还写过链表维护的，不过感觉都差不多，反正复杂度都是对的，肯定能过qwq。
Code
#include <bits/stdc++.h>
#define dprint printf
using namespace std;
typedef long long ll;
typedef pair<ll,int> pli;
const int MAXN = 100010;
int _w;
void noprint(...) {}
int fpow( int a, int b, int mod ) {
int c = 1;
while(b) {
if( b & 1 ) c = int(1LL * c * a % mod);
a = int(1LL * a * a % mod);
b >>= 1;
}
return c;
}
int inter( int l1, int r1, int l2, int r2 ) {
int l = max(l1, l2);
int r = min(r1, r2);
return max(r-l+1, 0);
}
int n, m, seed, vmax;
ll a[MAXN];
int rnd() {
const int MOD = 1e9+7;
int ret = seed;
seed = int((1LL * seed * 7 + 13) % MOD);
return ret;
}
pli b[MAXN];
int sz;
void prelude() {
sz = 1;
b[sz-1].first = a[1], b[sz-1].second = 0;
for( int i = 1; i <= n; ++i ) {
if( b[sz-1].first == a[i] ) {
++b[sz-1].second;
} else {
b[sz].first = a[i], b[sz].second = 1, ++sz;
}
}
}
pli c[MAXN];
int csz;
void append( ll x, int cnt ) {
if( csz && c[csz-1].first == x )
c[csz-1].second += cnt;
else
c[csz].first = x, c[csz].second = cnt, ++csz;
}
void solve1( int l, int r, int x ) {
int L = 0, R = 0;
csz = 0;
for( int i = 0; i < sz; ++i ) {
L = R+1;
R = L + b[i].second - 1;
ll num = b[i].first;
int cntl = inter(L, l-1, L, R);
int cnt = inter(L, R, l, r);
int cntr = inter(r+1, R, L, R);
if( cntl ) append(num, cntl);
if( cnt ) append(num+x, cnt);
if( cntr ) append(num, cntr);
}
sz = csz;
for( int i = 0; i < sz; ++i )
b[i] = c[i];
}
void solve2( int l, int r, int x ) {
int L = 0, R = 0;
csz = 0;
for( int i = 0; i < sz; ++i ) {
L = R+1;
R = L + b[i].second - 1;
ll num = b[i].first;
int cntl = inter(L, l-1, L, R);
int cnt = inter(L, R, l, r);
int cntr = inter(r+1, R, L, R);
if( cntl ) append(num, cntl);
if( cnt ) append(x, cnt);
if( cntr ) append(num, cntr);
}
sz = csz;
for( int i = 0; i < sz; ++i )
b[i] = c[i];
}
void solve3( int l, int r, int x ) {
int L = 0, R = 0;
csz = 0;
for( int i = 0; i < sz; ++i ) {
L = R+1;
R = L + b[i].second - 1;
int cnt = inter(L, R, l, r);
if( cnt ) c[csz++] = pli( b[i].first, cnt );
}
sort(c, c+csz);
L = R = 0;
for( int i = 0; i < csz; ++i ) {
L = R+1;
R = L + c[i].second - 1;
if( x >= L && x <= R )
return (void)printf( "%lld
", c[i].first );
}
return assert(0);
}
void solve4( int l, int r, int x, int y ) {
int L = 0, R = 0, ans = 0;
for( int i = 0; i < sz; ++i ) {
L = R+1;
R = L + b[i].second - 1;
int cnt = inter(L, R, l, r);
if( cnt ) ans = int((ans + 1LL * cnt * fpow( int(b[i].first % y), x, y )) % y);
}
printf( "%d
", ans );
}
void output_array() {
for( int i = 0; i < sz; ++i ) {
int t = b[i].second;
while( t-- )
dprint( "%lld ", b[i].first );
}
dprint("
");
}
int main() {
_w = scanf( "%d%d%d%d", &n, &m, &seed, &vmax );
// dprint("Init:
");
for( int i = 1; i <= n; ++i ) {
a[i] = rnd() % vmax + 1;
// dprint("%lld ", a[i]);
}
// dprint("
");
prelude();
for( int i = 1; i <= m; ++i ) {
int op, l, r, x, y;
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() % vmax + 1;
if( op == 4 )
y = rnd() % vmax + 1;
if( op == 1 ) solve1(l, r, x);
else if( op == 2 ) solve2(l, r, x);
else if( op == 3 ) solve3(l, r, x);
else if( op == 4 ) solve4(l, r, x, y);
// dprint("op = %d, l = %d, r = %d, x = %d, y = %d
", op, l, r, x, y);
// output_array();
}
return 0;
}