线段树模板整理

　　 线段树算是一种较为简单的中级数据结构了，线段树本身其实还是很简单的，就是一棵二叉树，每一个节点维护一块区间的信息，如果根节点是$[1,n]$，那么左儿子就是$[1,n/2]$,右儿子就是$[n/2+1,n]$，如此二分下去就是一棵线段树了，查找的时间复杂度是 $Oleft ( log_{2} N ight )$的，有图为证：(网上找的)

下面是某线段树模板题的代码：

#include <cstdio>
#include <algorithm>
using namespace std;

typedef long long LL;
static const int maxm=1e6+10;
LL tree[maxm],lazy[maxm],A[maxm],left[maxm],right[maxm];
int n,m;

void build(int num,int l,int r){
    left[num]=l;right[num]=r;
    int mid=(l+r)>>1;
    if(l==r){ tree[num]=A[l]; return; }
    build(num<<1,l,mid);
    build(num<<1|1,mid+1,r);
    tree[num]=tree[num<<1]+tree[num<<1|1];
}

void pushdown(int num){
    if(lazy[num]){
        int mid=(left[num]+right[num])>>1;
        tree[num<<1]+=(mid-left[num]+1)*lazy[num];
        tree[num<<1|1]+=(right[num]-mid)*lazy[num];
        lazy[num<<1]+=lazy[num];
        lazy[num<<1|1]+=lazy[num];
        lazy[num]=0;
    }
}

void update(int num,int l,int r,LL add){
    if(left[num]>=l&&right[num]<=r){
        tree[num]+=(right[num]-left[num]+1)*add;
        lazy[num]+=add;
        return;
    }
    if(left[num]>r||right[num]<l)return;
    pushdown(num);
    update(num<<1,l,r,add);
    update(num<<1|1,l,r,add);
    tree[num]=tree[num<<1]+tree[num<<1|1];
}

LL Query(int num,int l,int r){
    if(left[num]>=l&&right[num]<=r)return tree[num];
    if(left[num]>r||right[num]<l) return 0;
    LL ret=0;
    pushdown(num);
    ret+=Query(num<<1,l,r);
    ret+=Query(num<<1|1,l,r);
    return ret;
}

int main(){
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)scanf("%lld",&A[i]);
    build(1,1,n);
    for(int i=1;i<=m;i++){
        int f,x,y;LL add;
        scanf("%d",&f);
        switch(f){
            case 1:scanf("%d%d%lld",&x,&y,&add);update(1,x,y,add);break;
            case 2:scanf("%d%d",&x,&y);printf("%lld
",Query(1,x,y));break;
            default:printf("Orz %%%");break;
        }
    }

    return 0;
}

以下是维护序列的代码：(滋瓷乘法运算)

#include <cstdio>

typedef long long LL;

static const int maxm=100005;

LL A[maxm],pls[maxm<<2],mul[maxm<<2],tr[maxm<<2];
int left[maxm<<2],right[maxm<<2];
int n,m;
LL MOD;

int build(int id,int l,int r){
    left[id]=l;right[id]=r;mul[id]=1;
    if(l==r)return tr[id]=A[l]%MOD,0;
    int mid=(l+r)>>1;
    build(id<<1,l,mid);
    build(id<<1|1,mid+1,r);
    tr[id]=(tr[id<<1]+tr[id<<1|1])%MOD;
}

void pushdown(int id){
    int l=left[id];int r=right[id];int mid=(l+r)>>1;
    pls[id<<1]=(pls[id<<1]*mul[id]%MOD+pls[id])%MOD;
    pls[id<<1|1]=(pls[id<<1|1]*mul[id]%MOD+pls[id])%MOD;
    mul[id<<1]=(mul[id]*mul[id<<1])%MOD;
    mul[id<<1|1]=(mul[id]*mul[id<<1|1])%MOD;
    tr[id<<1]=(tr[id<<1]*mul[id]%MOD+pls[id]*(mid-l+1)%MOD)%MOD;
    tr[id<<1|1]=(tr[id<<1|1]*mul[id]%MOD+pls[id]*(r-mid)%MOD)%MOD;
    mul[id]=1;pls[id]=0;
}

void modify(int id,int l,int r,int c,int opt){
    if(left[id]>=l&&right[id]<=r){
        if(opt==1){
            mul[id]=(mul[id]*c)%MOD;
            pls[id]=(pls[id]*c)%MOD;
            tr[id]=(tr[id]*c)%MOD;
        }else if(opt==2){
            pls[id]=(pls[id]+c)%MOD;
            tr[id]=(tr[id]+(LL)c*(right[id]-left[id]+1)%MOD)%MOD;
        }
        return ;
    }
    if(right[id]<l||left[id]>r)return ;
    pushdown(id);
    modify(id<<1,l,r,c,opt);
    modify(id<<1|1,l,r,c,opt);
    tr[id]=(tr[id<<1]+tr[id<<1|1])%MOD;
}

LL Query(int id,int l,int r){
    if(left[id]>=l&&right[id]<=r)return tr[id]%MOD;
    if(left[id]>r||right[id]<l)return 0;
    pushdown(id);
    return (Query(id<<1,l,r)%MOD+Query(id<<1|1,l,r)%MOD)%MOD;
}

int main(){
    int opt,l,r,c;
    scanf("%d%lld",&n,&MOD);
    for(int i=1;i<=n;i++)scanf("%lld",&A[i]);
    scanf("%d",&m);

    build(1,1,n);
    
    while(m--){
        scanf("%d",&opt);
        if(opt!=3){
            scanf("%d%d%d",&l,&r,&c);
            modify(1,l,r,c,opt);
        }
        else scanf("%d%d",&l,&r),printf("%lld
",Query(1,l,r)%MOD);
    }

    return 0;
}