A Simple Problem with Integers(线段树区间更新复习，lazy数组的应用）-------------------蓝桥备战系列...-白红宇

A Simple Problem with Integers(线段树区间更新复习，lazy数组的应用）-------------------蓝桥备战系列...

阅读量：4622 次

发布时间：2019-06-09

本文共 2420 字，大约阅读时间需要 8 分钟。

You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is to ask for the sum of numbers in a given interval.

Input

The first line contains two numbers N and Q. 1 ≤ N,Q ≤ 100000.

The second line contains N numbers, the initial values of A1, A2, ... , AN. -1000000000 ≤ Ai ≤ 1000000000.

Each of the next Q lines represents an operation.

"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.

Output

You need to answer all Q commands in order. One answer in a line.

Sample Input

10 51 2 3 4 5 6 7 8 9 10Q 4 4Q 1 10Q 2 4C 3 6 3Q 2 4

Sample Output

Hint

The sums may exceed the range of 32-bit integers.

代码：

#include
    
     #include
     
      #include
      
       #include
       
        #include
        
         #include
         
          #include
          #include
           
            #include
            
             #include
             
              const int maxn=1e5+5;typedef long long ll;using namespace std;struct node{ ll l,r,sum;}tree[maxn<<2];ll lazy[maxn<<2];void pushup(int m){ tree[m].sum=tree[m<<1].sum+tree[m<<1|1].sum;}void pushdown(int m,int l){ if(lazy[m]!=0) { lazy[m<<1]+=lazy[m]; lazy[m<<1|1]+=lazy[m]; tree[m<<1].sum+=lazy[m]*(l-(l>>1)); tree[m<<1|1].sum+=lazy[m]*(l>>1); lazy[m]=0; }}void build(int m,int l,int r){ tree[m].l=l; tree[m].r=r; if(l==r) { scanf("%lld",&tree[m].sum); return; } int mid=(l+r)>>1; build(m<<1,l,mid); build(m<<1|1,mid+1,r); pushup(m);}void update(int m,int l,int r,int val){ if(tree[m].l==l&&tree[m].r==r) { lazy[m]+=val; tree[m].sum+=(ll)val*(r-l+1); return; } if(tree[m].l==tree[m].r) return; int mid=(tree[m].l+tree[m].r)>>1; pushdown(m,tree[m].r-tree[m].l+1); if(r<=mid) { update(m<<1,l,r,val); } else if(l>mid) { update(m<<1|1,l,r,val); } else { update(m<<1,l,mid,val); update(m<<1|1,mid+1,r,val); } pushup(m);}ll query(int m,int l,int r){ if(tree[m].l==l&&tree[m].r==r) { return tree[m].sum; } pushdown(m,tree[m].r-tree[m].l+1); int mid=(tree[m].l+tree[m].r)>>1; ll res=0; if(r<=mid) { res+=query(m<<1,l,r); } else if(l>mid) { res+=query(m<<1|1,l,r); } else { res+=(query(m<<1,l,mid)+query(m<<1|1,mid+1,r)); } return res; }int main(){ std::ios::sync_with_stdio(false); std::cin.tie(0); int n,m; cin>>n>>m; build(1,1,n); char op[2]; int l,r,val; for(int t=0;t