POJ 3468 A Simple Problem with Integers(线段树区间更新，模板题，求区间和)

#include <iostream>
#include <stdio.h>
#include <string.h>
#define lson rt<<1,L,mid
#define rson rt<<1|1,mid+1,R

using namespace std;
const int maxn=;
int n,q;
long long num[maxn];
struct Node{
    long long sum,add;
    bool lazy;
}tree[maxn<<];

void pushUp(int rt){
    tree[rt].sum=tree[rt<<].sum+tree[rt<<|].sum;
}
void pushDown(int rt,int m){
    if(tree[rt].lazy){
        tree[rt<<].add+=tree[rt].add;
        tree[rt<<|].add+=tree[rt].add;
        tree[rt<<].sum+=(m-m/)*tree[rt].add;  //是加上父亲的add
        tree[rt<<|].sum+=m/*tree[rt].add;
        tree[rt<<].lazy=tree[rt<<|].lazy=true;
        tree[rt].add=;
        tree[rt].lazy=false;
    }
}
void build(int rt,int L,int R){
    if(L==R){
        tree[rt].sum=num[L];
        tree[rt].add=;
        tree[rt].lazy=false;
        return ;
    }
    int mid=(L+R)>>;
    build(lson);
    build(rson);
    pushUp(rt);
}

void update(int rt,int L,int R,int l,int r,long long val){
    if(l<=L&&R<=r){
        tree[rt].sum+=(R-L+)*val;
        tree[rt].add+=val;
        tree[rt].lazy=true;
        return;
    }
    pushDown(rt,R-L+);
    int mid=(L+R)>>;
    if(l<=mid)
        update(lson,l,r,val);
    if(r>mid)
        update(rson,l,r,val);
    pushUp(rt);
}

long long query(int rt,int L,int R,int l,int r){
    long long ret=;
    if(l<=L&&R<=r){
        return tree[rt].sum;
    }
    pushDown(rt,R-L+);  //额，好吧，忘记这里也要初始化了
    int mid=(L+R)>>;
    if(l<=mid)
        ret+=query(lson,l,r);
    if(r>mid)
        ret+=query(rson,l,r);
    return ret;
}
int main()
{
    int a,b;
    long long c; //c用int也能AC，用long long是防止当(R-L+1)*c的时候超出int范围
    char str[];
    long long ans;
    scanf("%d%d",&n,&q);
    for(int i=;i<=n;i++)
        scanf("%I64d",&num[i]);
    build(,,n);
    for(int i=;i<=q;i++){
        scanf("%s",str);
        if(str[]=='C'){
            scanf("%d%d%I64d",&a,&b,&c);
            update(,,n,a,b,c);
        }
        else{
            scanf("%d%d",&a,&b);
            ans=query(,,n,a,b);
            printf("%I64d\n",ans);
        }

    }
    return ;
}