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  • 【poj3468】A Simple Problem with Integers

    题面

    You have N integers, A1, A2, … , AN. You need to deal with two kinds of operations.
    One type of operation is to add some given number to each number in a given interval.
    The other is to ask for the sum of numbers in a given interval.

    题解

    线段树模板

    #include<iostream>
    #include<algorithm>
    using namespace std;
    const int maxn = 100010;
    typedef long long LL;
    
    LL a[maxn];
    struct node{
        int l, r;
        LL val, addmark;
    }sgt[maxn<<2];
    void build(int p, int l, int r){
        sgt[p].l = l, sgt[p].r = r;
        if(l == r){
            sgt[p].val = a[l];
        }else{
            int m = (l+r)/2;
            build(p*2,l,m);
            build(p*2+1,m+1,r);
            sgt[p].val = sgt[p*2].val+sgt[p*2+1].val;
        }
    }
    void pushdown(int p){
        if(sgt[p].addmark != 0){
            LL t = sgt[p].addmark;
            sgt[p].addmark = 0;
            sgt[p*2].addmark += t;
            sgt[p*2+1].addmark += t;
            sgt[p*2].val += (sgt[p*2].r-sgt[p*2].l+1)*t;
            sgt[p*2+1].val += (sgt[p*2+1].r-sgt[p*2+1].l+1)*t;
        }
    }
    void add(int p, int l, int r, LL v){
        if(l <= sgt[p].l && sgt[p].r <= r){
            sgt[p].val += (sgt[p].r-sgt[p].l+1)*v;
            sgt[p].addmark += v;
            return ;
        }
        pushdown(p);
        int m = (sgt[p].l+sgt[p].r)/2;
        if(l <= m)add(p*2,l,r,v);
        if(r > m)add(p*2+1,l,r,v);
        sgt[p].val = sgt[p*2].val+sgt[p*2+1].val;
    }
    LL query(int p, int l, int r){
        if(l <= sgt[p].l && sgt[p].r <= r)return sgt[p].val;
        pushdown(p); //pushdown
        LL m = (sgt[p].l+sgt[p].r)/2, ans = 0;
        if(l <= m)ans += query(p*2,l,r);
        if(r > m)ans += query(p*2+1,l,r);
        return ans;
    }
    
    int main(){
        ios::sync_with_stdio(false);
        int n, m;
        cin>>n>>m;
        for(int i = 1; i <= n; i++)cin>>a[i];
        build(1,1,n);
        for(int i = 1; i <= m; i++){
            char ch;  cin>>ch;
            if(ch == 'Q'){
                LL x, y;  cin>>x>>y;
                cout<<query(1,x,y)<<"
    ";
            }else{
                LL x, y, z;  cin>>x>>y>>z;
                add(1,x,y,z);
            }
        }
        return 0;
    }

    关于线段树2*n空间表示法。
    推荐博客http://www.cppblog.com/MatoNo1/archive/2015/05/05/195857.html

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  • 原文地址:https://www.cnblogs.com/gwj1314/p/9444871.html
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