hdu4578 Transformation_hdu transformation-优快云博客

本文详细介绍了线段树的构建、下拉更新、区间更新及区间查询等核心操作，通过实例展示了如何利用线段树高效处理重复数据结构问题。重点强调了牛哥的巧妙算法，提供了解决复杂数据结构问题的新思路。

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转自 http://blog.youkuaiyun.com/ophunter_lcm/article/details/9878105

成名哥果然牛逼，这种题分类为简单线段树，又学了两个操作，现在看来，数据结构遇到牛逼的操作也只能缩短重复的东西。。。。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <vector>
using namespace std;
#define mod 10007
#define maxn 1100000
#define lson rt<<1,l,m
#define rson rt<<1|1,m+1,r
struct node
{
    int sum,multi,add;
    int lazy;
}tree[maxn<<2];

void build(int rt,int l,int r)
{
    int m=(l+r)>>1;
    tree[rt].multi=1;
    tree[rt].add=tree[rt].sum=tree[rt].lazy=0;
    if(l==r)
    {
        tree[rt].lazy=1;
        return ;
    }
    build(lson);
    build(rson);
}
void pushdown(int rt)
{
    if(tree[rt].lazy)
    {
        tree[rt<<1].lazy = tree[rt<<1|1].lazy = 1;
        tree[rt<<1].sum = tree[rt<<1|1].sum = tree[rt].sum;
        tree[rt<<1].add = tree[rt<<1|1].add = 0;
        tree[rt<<1].multi = tree[rt<<1|1].multi = 1;
        tree[rt].lazy = 0;
        return;
    }
    if(tree[rt].add>0)
    {
        if(tree[rt<<1].lazy)
        {
            tree[rt<<1].sum += tree[rt].add;
            tree[rt<<1].sum %= mod;
        }
        else
        {
            pushdown(rt<<1);
            tree[rt<<1].add += tree[rt].add;
            tree[rt<<1].add %= mod;
        }
        if(tree[rt<<1|1].lazy)
        {
            tree[rt<<1|1].sum += tree[rt].add;
            tree[rt<<1|1].sum %= mod;
        }
        else
        {
            pushdown(rt<<1|1);
            tree[rt<<1|1].add += tree[rt].add;
            tree[rt<<1|1].add %= mod;
        }
        tree[rt].add = 0;
    }
    if(tree[rt].multi > 1)
    {
        if(tree[rt<<1].lazy)
        {
            tree[rt<<1].sum *= tree[rt].multi;
            tree[rt<<1].sum %= mod;
        }
        else
        {
            pushdown(rt<<1);
            tree[rt<<1].multi *= tree[rt].multi;
            tree[rt<<1].multi %= mod;
        }
        if(tree[rt<<1|1].lazy)
        {
            tree[rt<<1|1].sum *= tree[rt].multi;
            tree[rt<<1|1].sum %= mod;
        }
        else
        {
            pushdown(rt<<1|1);
            tree[rt<<1|1].multi *= tree[rt].multi;
            tree[rt<<1|1].multi %= mod;
        }
        tree[rt].multi = 1;
    }
}
void update(int rt,int l,int r,int x,int y,int val,int op)
{
    if( l >= x && y >= r)
    {
        if(op == 3)
        {
            tree[rt].lazy = 1;
            tree[rt].sum = val;
            tree[rt].add = 0;
            tree[rt].multi = 1;
        }
        else
        {
            if(tree[rt].lazy)
            {
                if(op == 1)
                    tree[rt].sum += val;
                else
                    tree[rt].sum *= val;
                tree[rt].sum %= mod;
            }
            else
            {
                pushdown(rt);
                if(op == 1)
                {
                    tree[rt].add += val;
                    tree[rt].add %= mod;
                }
                else
                {
                    tree[rt].multi *= val;
                    tree[rt].multi %= mod;
                }
            }
        }
        return;
    }
    pushdown(rt);
    int m=(l+r)>>1;
    if(x<=m)
    {
        update(lson,x,y,val,op);
    }
    if(y>m)
    {
        update(rson,x,y,val,op);
    }
}
int query(int num,int s,int e,int l,int r,int p)
{
    if(s >= l && r >= e)
    {
        if(tree[num].lazy)
        {
            long long ret = 1;
            while(p --)
            {
                ret *= tree[num].sum;
                ret %= mod;
            }
            ret *= (e - s + 1);
            ret %= mod;
            return (int)ret;
        }
    }
    pushdown(num);
    int mid = (s + e)>>1;
    int tmp=0;
    if(l <= mid)
    {
        tmp+=query(num<<1,s,mid,l,r,p);
    }
    if(r > mid)
    {
        tmp+=query(num<<1|1,mid + 1,e,l,r,p);
    }
    return tmp;
}
int main()
{
    int n,m;
    int op,a,b,c;
    while(scanf("%d%d",&n,&m) != EOF)
    {
        if(n == 0 && m == 0)
            break;
        build(1,1,n);
        while(m --)
        {
            scanf("%d%d%d%d",&op,&a,&b,&c);
            if(op <= 3)
                update(1,1,n,a,b,c,op);
            else
                printf("%d\n",query(1,1,n,a,b,c)%mod);
        }
    }
    return 0;
}