POJ - 3468（水题）树状数组区间修改区间查询模板

U have N integers, A1, A2, ... , AN
"C a b c" means adding c to each of Aa, Aa+1, ... , Ab. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of Aa, Aa+1, ... , Ab.

#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
typedef long long ll;
const ll maxn=100005;
ll delt[maxn],sum[maxn],n,m;
ll lowbit(ll x) {return x&(-x);}
void update(ll *a,ll pos,ll val)
{
    while(pos<=n+1)
    {
        a[pos]+=val;
        pos+=lowbit(pos);
    }
}
ll query(ll *a,ll pos)
{
    ll ans=0;
    while(pos>0)
    {
        ans+=a[pos];
        pos-=lowbit(pos);
    }
    return ans;
}
int main()
{
    #define int ll
    scanf("%lld%lld",&n,&m);
    int x=0,y=0;
    for(int i=1;i<=n;i++)
    {
        scanf("%lld",&x);
        update(delt,i,x-y);
        update(sum,i,(i-1)*(x-y));
        y=x;
    }
    //scanf("%d",&m);
    for(int i=1;i<=m;i++)
    {
        int x,y,z;
        char k[5];
        scanf("%s%lld%lld",k,&x,&y);
        if(k[0]=='C')
        {
            scanf("%lld",&z);
            update(delt,x,z);
            update(delt,y+1,-z);
            update(sum,x,z*(x-1));
            update(sum,y+1,-z*y);
        }
        else
        {
            int sum1=(x-1)*query(delt,x-1)-query(sum,x-1);
            int sum2=y*query(delt,y)-query(sum,y);
            printf("%lld\n",sum2-sum1);
        }
    }
    return 0;
} 

与线段树对比

#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <string>
#include <vector>
#include <map>
#include <queue>
#include <stack>

using namespace std;
typedef long long ll;
int n,m;
const int maxn=400000+100;
ll a[maxn],delt[maxn];

void pushup(int rt)
{
    a[rt]=a[rt<<1]+a[rt<<1|1];
}

void build(int rt,int l,int r)
{
    if(l==r)
    {
        scanf("%lld",&a[rt]);
        return ;
    }
    int mid=(l+r)>>1;
    build(rt<<1,l,mid);
    build(rt<<1|1,mid+1,r);
    pushup(rt);
}

void pushdown(int rt,int l,int r)
{
    if(delt[rt])
    {
        delt[rt<<1]+=delt[rt];
        delt[rt<<1|1]+=delt[rt];
        int mid=(l+r)>>1;
        a[rt<<1]+=(ll)(mid-l+1)*delt[rt];
        a[rt<<1|1]+=(ll)(r-mid)*delt[rt];
        delt[rt]=0;
    }
}

void update(int rt,int l,int r,int x,int y,ll val)
{
    if(x<=l&&y>=r)
    {
        delt[rt]+=val;
        a[rt]+=val*(ll)(r-l+1);
        return ;
    }
    pushdown(rt,l,r);
    int mid=(l+r)>>1;
    if(mid>=x)
        update(rt<<1,l,mid,x,y,val);
    if(mid<y)
        update(rt<<1|1,mid+1,r,x,y,val);
    pushup(rt);
}

ll query(int rt,int l,int r,int x,int y)
{
    if(l>=x&&r<=y)
    {
        return a[rt];
    }
    pushdown(rt,l,r);
    int mid=(l+r)>>1;
    ll ans=0;
    if(mid>=x)
        ans+=query(rt<<1,l,mid,x,y);
    if(mid<y)
        ans+=query(rt<<1|1,mid+1,r,x,y);
    return ans;
}


int main()
{
   // freopen("in.txt","r",stdin);
    while(~scanf("%d%d",&n,&m))
    {
        memset(delt,0,sizeof delt);
        build(1,1,n);
        while(m--)
        {
            char op[5];
            int x,y;
            ll val;
            scanf("%s",&op);
            if(op[0]=='Q')
            {
                scanf("%d%d",&x,&y);
                ll ans=query(1,1,n,x,y);
                printf("%lld\n",ans);
            }
            else
            {
                scanf("%d%d%lld",&x,&y,&val);
                update(1,1,n,x,y,val);
            }
        }
    }
    return 0;
}