Codeforces 455 D. serega and fun

$> Codeforces \ Space 455 D .? Serega? And? Fun <$

Theme:
You have a sequence with a length of \ (n \). You must support two operations.

1. Given \ (L, r \), convert \ (A _ {I + 1} \) to \ (a_ I \), \ (a_l \) to \ (a_r, I \ in [L, R )\)

2. Returns \ (L, R, X \), and calculates the number of occurrences of \ (x \) in \ ([L, R] \).

\ (1 \ Leq n, m \ Leq 10 ^ 5 \)

Solution: Song

/*program by mangoyang*/#pragma GCC optimize("Ofast","O3", "inline","-ffast-math")#pragma GCC target("avx,sse2,sse3,sse4,mmx")#include<bits/stdc++.h>#define inf (0x7f7f7f7f)#define Max(a, b) ((a) > (b) ? (a) : (b))#define Min(a, b) ((a) < (b) ? (a) : (b))typedef long long ll;using namespace std;     inline char read() {    static const int IN_LEN = 200000;    static char buf[IN_LEN], *s, *t;    return (s == t ? t = (s = buf) + fread(buf, 1, IN_LEN, stdin), (s == t ? -1 : *s++) : *s++);}template<class T>inline void read(T &x) {    static bool iosig;    static char c;    for (iosig = false, c = read(); !isdigit(c); c = read()) {        if (c == '-') iosig = true;        if (c == -1) return;    }    for (x = 0; isdigit(c); c = read()) x = ((x + (x << 2)) << 1) + (c ^ '0');    if (iosig) x = -x;}const int OUT_LEN = 500000;char obuf[OUT_LEN], *ooh = obuf;inline void print(char c) {    if (ooh == obuf + OUT_LEN) fwrite(obuf, 1, OUT_LEN, stdout), ooh = obuf;    *ooh++ = c;}template<class T>inline void print(T x) {    static int buf[30], cnt;    if (x == 0) print('0');    else {        if (x < 0) print('-'), x = -x;        for (cnt = 0; x; x /= 10) buf[++cnt] = x % 10 + 48;        while (cnt) print((char)buf[cnt--]);    }}inline void flush() { fwrite(obuf, 1, ooh - obuf, stdout); }const int N = 100005;int a[N], s[N], n, m, lastans;int main(){    read(n);    for(register int i = 1; i <= n; ++i) read(a[i]), s[a[i]]++;    register int *p; read(m);    for(register int i = 1, j, op, x, y, z, tmp, res; i <= m; ++i){        read(op), read(x), read(y);        x = (x + lastans - 1) % n + 1, y = (y + lastans - 1) % n + 1;        if(x > y) swap(x, y);        if(op & 1){            p = a + y, tmp = *p, j = y - x;            while(j) *p = *(p-1), --p, --j; *p = tmp;        }        else{            read(z), z = (z + lastans - 1) % n + 1, res = 0;            if(y - x + 1<= n / 2){                p = a + x - 1;                for(j = x; j + 8 <= y; j += 8) {                    res += (*(++p) == z), res += (*(++p) == z);                    res += (*(++p) == z), res += (*(++p) == z);                    res += (*(++p) == z), res += (*(++p) == z);                    res += (*(++p) == z), res += (*(++p) == z);                }                for(; j <= y; j++) res += (*(++p) == z);                print(lastans = res), print('\n');            }            else{                p = a;                for(j = 1; j + 8 < x; j += 8){                    res += (*(++p) == z), res += (*(++p) == z);                    res += (*(++p) == z), res += (*(++p) == z);                    res += (*(++p) == z), res += (*(++p) == z);                    res += (*(++p) == z), res += (*(++p) == z);                }                for(; j < x; j++) res += (*(++p) == z);                p = a + y;                for(j = y + 1; j + 8 <= n; j += 8){                    res += (*(++p) == z), res += (*(++p) == z);                    res += (*(++p) == z), res += (*(++p) == z);                    res += (*(++p) == z), res += (*(++p) == z);                    res += (*(++p) == z), res += (*(++p) == z);                }                for(; j <= n; j++) res += (*(++p) == z);                print(lastans = s[z] - res), print('\n');            }        }    }    return flush(), 0;}

Codeforces 455 D. serega and fun