「分塊系列」數列分塊入門7 解題報告

時間 2020-01-08

標籤分塊系列數列入門解題報告欄目應用數學简体版

原文原文鏈接

數列分塊入門7

題意歸納

區間乘法，區間加法，單點詢問。html

寫在前面

寫過線段樹模板2的童鞋應該很清楚了吧QAQc++

因爲*與Markdown衝突，因此用×代替o(*￣︶￣*)ospa

正題

咱們把一個數表示爲 a[i] × tg2[b[i]] + tg1[b[i]]。tg2表示乘法標記，tg1表示加法標記。code

對於不完整的塊，直接 a[i] = a[i] × tg2[b[i]] + tg1[b[i]] 將這個塊的全部元素都還原，也就是將該塊的標記下傳。htm

對於完整的塊blog

乘法：( a[i] × tg2[b[i]] + tg1[b[i]] ) × c = a[i] × (tg2[b[i]] × c) + (tg1[b[i]] × c) 也就是說，將tg一、tg2都乘c就能夠了get

加法：( a[i] × tg2[b[i]] + tg1[b[i]] ) + c = a[i] × tg2[b[i]] + (tg1[b[i]] + c) 也就是將tg1加上cit

而後就很清楚了ヾ(ｏ･ω･)ﾉ入門

代碼

#include<bits/stdc++.h>
using namespace std;
#define MAXN 100005
#define mod(x) (1ll * x) % 10007

int n, d;
int a[MAXN], b[MAXN], tg1[500], tg2[500];

inline void Push( int wh ){
    for ( int i = ( wh - 1 ) * d + 1; i <= wh * d; ++i ) a[i] = mod( 1ll * a[i] * tg2[wh] + tg1[wh] );
    tg1[wh] = 0; tg2[wh] = 1;
}

void Add( int l, int r, int c ){
    if ( b[l] == b[r] ){
        Push(b[l]);
        for ( int i = l; i <= r; ++i ) a[i] = mod( a[i] + c );
        return;
    }
    Push(b[l]);
    for ( int i = l; b[i] == b[l]; ++i ) a[i] = mod( a[i] + c );
    Push(b[r]);
    for ( int i = r; b[i] == b[r]; --i ) a[i] = mod( a[i] + c );
    for ( int i = b[l] + 1; i <= b[r] - 1; ++i ) tg1[i] = mod( tg1[i] + c );
}

void Mul( int l, int r, int c ){
    if ( b[l] == b[r] ){
        Push(b[l]);
        for ( int i = l; i <= r; ++i ) a[i] = mod( a[i] * c );
        return;
    }
    Push(b[l]);
    for ( int i = l; b[i] == b[l]; ++i ) a[i] = mod( a[i] * c );
    Push(b[r]);
    for ( int i = r; b[i] == b[r]; --i ) a[i] = mod( a[i] * c );
    for ( int i = b[l] + 1; i <= b[r] - 1; ++i ) tg1[i] = mod( tg1[i] * c ), tg2[i] = mod( tg2[i] * c );
}

int main(){
    scanf( "%d", &n );
    d = sqrt(n);
    for ( int i = 1; i <= n; ++i ){
        scanf( "%d", &a[i] );
        b[i] = ( i - 1 ) / d + 1;
    }
    for ( int i = 1; i <= b[n]; ++i ) tg1[i] = 0, tg2[i] = 1;
    for ( int i = 1; i <= n; ++i ){
        int opt, l, r, c;
        scanf( "%d%d%d%d", &opt, &l, &r, &c );
        if ( opt == 0 ) Add( l, r, c );
        if ( opt == 1 ) Mul( l, r, c );
        if ( opt == 2 ) printf( "%d\n", mod(a[r] * tg2[b[r]] + tg1[b[r]]) );
    }
    return 0;
}