数列分块入门7
题意概括
区间乘法，区间加法，单点询问。
写在前面
写过线段树模板2的童鞋应该很清楚了吧QAQ
由于*与Markdown冲突，所以用×代替o(*￣︶￣*)o
正题
我们把一个数表示为 a[i] × tg2[b[i]] + tg1[b[i]]。tg2表示乘法标记，tg1表示加法标记。
对于不完整的块，直接 a[i] = a[i] × tg2[b[i]] + tg1[b[i]] 将这个块的所有元素都还原，也就是将该块的标记下传。
对于完整的块
乘法：( a[i] × tg2[b[i]] + tg1[b[i]] ) × c = a[i] × (tg2[b[i]] × c) + (tg1[b[i]] × c) 也就是说，将tg1、tg2都乘c就可以了
加法：( a[i] × tg2[b[i]] + tg1[b[i]] ) + c = a[i] × tg2[b[i]] + (tg1[b[i]] + c) 也就是将tg1加上c
然后就很清楚了ヾ(ｏ･ω･)ﾉ
代码
#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
", mod(a[r] * tg2[b[r]] + tg1[b[r]]) );
}
return 0;
}
总结
有多种操作时可以借助代数来分析~(＾ω＾)
数列分块系列目录
数列分块入门7 <-