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洛谷P4168

思路

这道题和 数列分块入门 9 蜜汁相似QAQ。
请自行参照我的 数列分块入门 9 解题报告
这里仅放上代码QAQ——

代码

#include<bits/stdc++.h>
using namespace std;
#define MAXN 40005

int n, m, T;
int a[MAXN], b[MAXN], c[MAXN];
int d, f[2000][2000];
int s[MAXN];
vector<int> p[MAXN];

int Count( int l, int r, int x ){
    return upper_bound( p[x].begin(), p[x].end(), r ) - lower_bound( p[x].begin(), p[x].end(), l );
}

int Get( int l, int r ){
    if ( b[l] == b[r] ){
        int ans1(0), ans2(0);
        for ( int i = l; i <= r; ++i ){
            int t(Count( l, r, a[i] ));
            if ( t > ans2 ) ans1 = a[i], ans2 = t;
            if ( t == ans2 ) ans1 = min( ans1, a[i] );
        }
        return ans1;
    }
    int ans1(f[b[l] + 1][b[r] - 1]), ans2(Count( l, r, ans1 ));
    for ( int i = l; b[l] == b[i]; ++i ){
        int t(Count( l, r, a[i] ));
        if ( t > ans2 ) ans1 = a[i], ans2 = t;
        if ( t == ans2 ) ans1 = min( ans1, a[i] );
    }
    for ( int i = r; b[r] == b[i]; --i ){
        int t(Count( l, r, a[i] ));
        if ( t > ans2 ) ans1 = a[i], ans2 = t;
        if ( t == ans2 ) ans1 = min( ans1, a[i] );
    }
    return ans1;
}

int main(){
    scanf( "%d%d", &n, &T );
    d = 0;
    while( ( 1 << d ) <= n ) d++;
    d = (int)( n / sqrt( 2 * T * d ) );
    for ( int i = 1; i <= n; ++i ){
        scanf( "%d", &a[i] ); c[i] = a[i]; b[i] = ( i - 1 ) / d + 1;
    }
    sort( c + 1, c + n + 1 );
    m = unique( c + 1, c + n + 1 ) - c - 1;
    for ( int i = 1; i <= n; ++i ) a[i] = lower_bound( c + 1, c + m + 1, a[i] ) - c;
    for ( int i = 1; i <= n; ++i ) p[a[i]].push_back(i);

    for ( int i = 1; i <= b[n]; ++i ){
        memset( s, 0, sizeof s );
        int ans1(0), ans2(0);
        for ( int j = ( i - 1 ) * d + 1; j <= n; ++j ){
            s[a[j]]++;
            if ( s[a[j]] == ans2 ) ans1 = min( ans1, a[j] );
            if ( s[a[j]] > ans2 ) ans1 = a[j], ans2 = s[a[j]];
            if ( b[j + 1] != b[j] ) f[i][b[j]] = ans1;
        }
    }

    int x(0);
    while( T-- ){
        int l, r;
        scanf( "%d%d", &l, &r );
        l = ( l + x - 1 ) % n + 1; r = ( r + x - 1 ) % n + 1;
        int t(min( l, r )); r = max( l, r ); l = t;
        printf( "%d\n", x = c[Get( l, r )] );
    }
    return 0;
}