POJ 3468（线段树区间修改+区间求和）

题目：给个n给个m，给n个数，接下来m个指令，有两种指令，第一种让输出区间a到b的和，第二种把a到b的数改成c

线段树板子题，开始写的时候不知道有延迟标记，然后疯狂T

#include <iostream>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <string>
#include <map>
#include <iomanip>
#include <algorithm>
#include <queue>
#include <stack>
#include <set>
#include <vector>
//const int maxn = 1e5+5;
#define ll long long
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}

#define MAX INT_MAX
#define FOR(i,a,b) for( int i = a;i <= b;++i)
#define bug cout<<"--------------"<<endl
using namespace std;
ll a[110000];
struct SegmentTree
{
    ll l,r;
    ll sum;
    ll add;
    #define l(x) tree[x].l
    #define r(x) tree[x].r
    #define sum(x) tree[x].sum
    #define add(x) tree[x].add
}tree[110000*4];
void build(ll p,ll l,ll r)
{
    l(p) = l,r(p)=r;
    //=====================================
    if(l == r) {sum(p)=a[l] ;return ;}
    ll mid=(l+r)/2;
    build(p*2, l,mid);
    build(p*2+1 , mid+1,r);
    sum(p)+=sum(p*2)+sum(p*2+1);
}
void spread(ll p)
{
    if(add(p))
    {
        //====================================================
        sum(p*2)+=add(p) * (r(p*2)-l(p*2)+1);
        sum(p*2+1)+= add(p) * (r(p*2+1)-l(p*2+1)+1);
        add(p*2)+=add(p);
        add(p*2+1)+=add(p);
        add(p)=0;
    }
}
void change(ll p,ll l,ll r,ll d )
{
    if(l<=l(p) && r>=r(p))
    {
        sum(p)+=(ll  )d * (r(p)-l(p)+1);
        add(p) +=d;
        return ;
    }
    spread(p);
    ll mid = (l(p)+r(p))/2;
    if(l<=mid) change(p*2 , l, r, d);
    if(r>mid) change(p*2+1,l,r,d);
    sum(p) = sum(p*2)+sum(p*2+1);
}
ll ask(ll p,ll l,ll r)
{
    if(l<=l(p) && r>=r(p) ) return sum(p);
    spread(p);
    ll mid = (l(p)+r(p))/2;
    ll val=0;
    if(l<=mid) val+=ask(p*2,l,r);
    if(r>mid) val+=ask(p*2+1,l,r);
    return val;
}

int main()
{
    ll n,m;
    cin>>n>>m;
    FOR(i,1,n) cin>>a[i];
    build(1,1,n);
    char c;
    while (m--)
    {
        char op[2];
        int x, y, z;
        scanf("%s%d%d", op, &x, &y);
        if (op[0] == 'C')
        {
            scanf("%d", &z);
            change(1, x, y, z);
        }
        else
            printf("%lld
", ask(1, x, y));
    }
}