树状数组初步

树状数组

FUCTION:

(begin{cases}text{点单修改&&区间求和} text{区间修改&&单点查询} text{区间修改&&区间查询}end{cases})

原理：定义c[i],表示以i为结尾的前lowbit(i)个数(原数列)的和(后缀和)

维护c[i],根据上图,从底至上(二进制增加拼凑)更新code:

inline void update(int pos,int val){
 for(int i=pos;i<=n;i+=lowbit(i))c[i]+=val;
}

APP1：单点修改&&区间查询

区间查询：前缀和的思想，query(l,r)=sum(r)-sum(l-1)

code:query(实质是根据上图的自顶向底(二进制拆分))

inline void query(int pos){
    int ans=0;
    for(int i=pos;i;i-=lowbit(i))ans+=c[i];
    return ans;
}//调用query(r)-query(l-1)

APP2：区间修改&&单点查询

solution：维护差分数组d[i],根据d数组维护c[i],区间修改转化成单点修改(在端点处修改),单点查询转化成区间求和(差分数组的前缀和)

code:

update(l,val);
update(r+1,-val);//[l,r]+val
query(pos)//单点查询

APP3:区间修改&&区间查询

(sumlimits_{i=1}^pa_i=sumlimits_{i=1}^psumlimits_{j=1}^i d[j]=sumlimits_{i=1}^p(p+i-1)d[i]=(p+1)sumlimits_{i=1}^pd[i]-sumlimits_{i=1}^p d[i]times i)
维护(c[i])为(d[i]),(ci[i])为(d[i]times i)
修改：同上APP2
查询：((p+1)times sum_1-sum_2)
code:

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
const int MAXX=101000;
long long a[MAXX],d[MAXX],c[MAXX],ci[MAXX],di[MAXX];
int n,q;
inline long long lowbit(long long x){
 return x&-x;
}
inline void add(long long pos,long long val){
    for(long long i=pos;i<=n;i+=lowbit(i)){
     c[i]+=val;
     ci[i]+=val*pos;
    }
}
inline void update(long long l,long long r,long long val){
 add(l,val);
 add(r+1,-val);
}
inline long long query(long long pos){
 long long ans1=0;
 long long ans2=0;
   for(long long i=pos;i;i-=lowbit(i)){
     ans1+=c[i];
     ans2+=ci[i];
   }
   return ans1*(pos+1)-ans2;
}
int main(){
 scanf("%d%d",&n,&q);
 for(long long i=1;i<=n;++i){
  scanf("%lld",&a[i]);
     add(i,a[i]-a[i-1]);
 }
    for(long long i=1;i<=q;++i){
     int opt,l,r;
     long long val;
        scanf("%d%d%d",&opt,&l,&r);
        if(opt==1){
         scanf("%lld",&val);
         update(l,r,val);
        }
        else printf("%lldn",query(r)-query(l-1));
    }
    return 0;
}