A Simple Problem with Integers POJ

题目链接：
POJ-3468

大意：
区间修改，区间求和。线段树模板题。

线段树与树状数组比较：
线段树的功能更强大，但树状数组的常数较小。

#include <iostream>#include <cstring>#include <string>#include <cmath>using namespace std;typedef long long ll;#define mem(s,t) memset(s,t,sizeof(s))#define D(v) cout<<#v<<" "<<v<<endl#define inf 0id3f3f3f3f#define pb push_back#define  pii pair<int,int>//#define LOCALconst int MAXN =1e5+10;int n,a[MAXN],q;struct node{    ll lazy,sum;    int l,r;    void update(ll val){        sum+=1LL*(r-l+1)*val;        lazy+=val;    }}tree[4*MAXN];void push_up(int id){    tree[id].sum=tree[id<<1].sum+tree[id<<1|1].sum;}void push_down(int id){    int lazyval=tree[id].lazy;    if(lazyval){        tree[id<<1].update(lazyval);        tree[id<<1|1].update(lazyval);        tree[id].lazy=0;    }}void build(int id,int l,int r){    tree[id].l=l,tree[id].r=r;    tree[id].sum=tree[id].lazy=0;    if(l==r){        tree[id].sum=a[l];    }else {        int mid=(l+r)/2;        build(id<<1,l,mid);        build(id<<1|1,mid+1,r);        push_up(id);    }}void update(int id,int l,int r,int val){    int L=tree[id].l,R=tree[id].r;    if(l<=L&&R<=r){        tree[id].update(val);    }else {        push_down(id);        int mid=(L+R)/2;        if(mid>=l) update(id<<1,l,r,val);        if(r>mid) update(id<<1|1,l,r,val);        push_up(id);    }}ll query(int id,int l,int r){    ll ans=0;    int L=tree[id].l,R=tree[id].r;    if(l<=L&&R<=r){        return tree[id].sum;    }else {        push_down(id);        int mid=(L+R)/2;        if(mid>=l) ans+=query(id<<1,l,r);        if(r>mid) ans+=query(id<<1|1,l,r);        push_up(id);    }    return ans;}int main(){    scanf("%d%d",&n,&q);    for(int i=1;i<=n;i++){        scanf("%d",&a[i]);    }    build(1,1,n);//build(1,l,r)    for(int i=1;i<=q;i++){        int l,r,val;        char ch[2];        scanf("%s",ch);        if(ch[0]=='Q'){            scanf("%d%d",&l,&r);            printf("%lld\n",query(1,l,r));        }else {            scanf("%d%d%d",&l,&r,&val);            update(1,l,r,val);        }    }    return 0;}