(区间更新lazy)POJ3468 A Simple Problem with Integers

传送门：(区间更新lazy)POJ3468 A Simple Problem with Integers

模板题。

代码：

#include<iostream>
#include<cstdio>
using namespace std;
const int maxn=1e5+10;
int a[maxn];
typedef long long ll;

struct Tree{
    int l,r;
    ll sum,lazy;
    void update(int x){
        lazy+=x;
        sum+=1LL*(r-l+1)*x;
    }
}tree[maxn<<2];
int n,q;

void push_up(int x){
    tree[x].sum=tree[x<<1].sum+tree[x<<1|1].sum;
}

void push_down(int x){
    int lazyval=tree[x].lazy;
    if(lazyval){
        tree[x<<1].update(lazyval);
        tree[x<<1|1].update(lazyval);
        tree[x].lazy=0;
    }
}

void build(int x,int l,int r){
    tree[x].l=l,tree[x].r=r;
    tree[x].sum=tree[x].lazy=0;
    if(l==r){
        tree[x].sum=a[l];
        return ;
    }
    int mid=l+((r-l)>>1);
    build(x<<1,l,mid);
    build(x<<1|1,mid+1,r);
    push_up(x);
}

void update(int x,int l,int r,int val){
    int L=tree[x].l,R=tree[x].r;
    if(l<=L&&R<=r){
        tree[x].update(val);
        return ;
    }
    push_down(x);
    int mid=L+((R-L)>>1);
    if(mid>=l) update(x<<1,l,r,val);
    if(mid<r) update(x<<1|1,l,r,val);
    push_up(x);
}

ll query(int x,int l,int r){
    int L=tree[x].l,R=tree[x].r;
    if(l<=L&&R<=r) return tree[x].sum;
    push_down(x);
    ll ans=0;
    int mid=L+((R-L)>>1);
    if(mid>=l) ans+=query(x<<1,l,r);
    if(mid<r) ans+=query(x<<1|1,l,r);
    push_up(x);
    return ans;
}


int main(){
    scanf("%d%d",&n,&q);
    for(int i=1;i<=n;i++){
        scanf("%d",&a[i]);
    }
    build(1,1,n);
    char op;
    int a,b;
    ll c;
    while(q--){
        scanf(" %c",&op);
        if(op=='Q'){
            scanf("%d%d",&a,&b);
            printf("%lld\n",query(1,a,b));
        }else{
            scanf("%d%d%lld",&a,&b,&c);
            update(1,a,b,c);
        }
    }
    return 0;
}