Codevs 1080~1082 Line Segment Tree Exercise series (template)

Source: Internet
Author: User

Title Description Description
Give you the number of n, there are two operations:
1: Add x for all numbers of interval [a, b]
2: Ask for the number of intervals [A, b] and.

Enter a description input Description
The first line is a positive integer n, and the next n rows n integers,
Then the next positive integer q, each line represents the number of operations,
If the first number is 1, followed by 3 positive integers,
Indicates that each number in the interval [a, b] is increased by x, if it is 2,
Represents the number of the and of the action 2 ask interval [A, b].
Pascal player please do not use READLN to read in

outputs description output Description
One answer for each query output line

sample input to sample
3
1
2
3
2
1 2 3 2
2 2 3

sample output Sample outputs
9

Data Size & Hint
Data range
1<=n<=200000
1<=q<=200000

The problem: The three problems with the same code to change the main function can be AC (this is clearly a problem ok ╭ (╯^╰) ╮), the simplest segment tree template.

The code is as follows:

#include <cstdio> #include <cstring> #include <iostream> using namespace std;
typedef long Long LL;
const LL maxn=233333;
int XL[MAXN];
    struct JGT {//int l,r;
ll Sum,add;
}HAH[MAXN*4];
    void build (int p,int l,int R)//build a line segment tree 233 {hah[p].l=l;
    Hah[p].r=r;
        if (l==r) {hah[p].sum=xl[l];
    return;
    } int mid= (L+R)/2;
    Build (P*2,l,mid);
Build (p*2+1,mid+1,r);//Line tree is a binary tree 233 hah[p].sum=hah[p*2].sum+hah[p*2+1].sum;
        } void spread (int p)//to devolve the tag {if (Hah[p].add) {hah[p*2].sum+= (hah[p*2].r-hah[p*2].l+1) *hah[p].add;
        hah[p*2+1].sum+= (hah[p*2+1].r-hah[p*2+1].l+1) *hah[p].add;
        Hah[p*2].add+=hah[p].add;
        Hah[p*2+1].add+=hah[p].add;
    Hah[p].add=0;
    }} ll ask (int p,int l,int R)//interval query {if (l<=hah[p].l&&hah[p].r<=r) {return hah[p].sum;
    } spread (P);//decentralization marks ll ans=0;
    int mid= (HAH[P].R+HAH[P].L)/2;
    if (l<=mid) ans+=ask (p*2,l,r); if (mId<r) Ans+=ask (p*2+1,l,r);
return ans;
    } void Change (int p,int l,int r,ll x)//For each modification we make a tag that is evaluated only when needed (L&LT;=HAH[P].L&AMP;&AMP;HAH[P].R&LT;=R)
        {hah[p].sum+= (LL) (hah[p].r-hah[p].l+1) *x;
        Hah[p].add+=x;
    return;
    } spread (P);//decentralization Mark int mid= (HAH[P].R+HAH[P].L)/2;
    if (l<=mid) change (p*2,l,r,x);
if (mid<r) change (p*2+1,l,r,x);//Continue to modify child nodes hah[p].sum=hah[p*2].sum+hah[p*2+1].sum;
    } int main () {ll n,q;
    scanf ("%lld", &n);
    for (int i=1;i<=n;i++) scanf ("%d", &xl[i]);
    Build (1,1,n);//Achievements scanf ("%lld", &q);
        for (int i=1;i<=q;i++) {int ss;
        scanf ("%d", &AMP;SS);
            if (ss==1) {int A, B;
            ll X;
            scanf ("%d%d%lld", &a,&b,&x);
            Change (1,a,b,x);//Modify} if (ss==2) {int A, B;
            scanf ("%d%d", &a,&b);
        printf ("%lld\n", Ask (1,a,b));//Query}} return 0; }

Contact Us

The content source of this page is from Internet, which doesn't represent Alibaba Cloud's opinion; products and services mentioned on that page don't have any relationship with Alibaba Cloud. If the content of the page makes you feel confusing, please write us an email, we will handle the problem within 5 days after receiving your email.

If you find any instances of plagiarism from the community, please send an email to: info-contact@alibabacloud.com and provide relevant evidence. A staff member will contact you within 5 working days.

A Free Trial That Lets You Build Big!

Start building with 50+ products and up to 12 months usage for Elastic Compute Service

  • Sales Support

    1 on 1 presale consultation

  • After-Sales Support

    24/7 Technical Support 6 Free Tickets per Quarter Faster Response

  • Alibaba Cloud offers highly flexible support services tailored to meet your exact needs.