牛客15175 Tree Recovery 樹狀數組區間操作求和模板題

題目鏈接：牛客15175 Tree Recovery

題目描述

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.

輸入描述

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.

輸出描述

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

題解

建議若想學習樹狀數組區間操作+區間和的同學，可以先去學習樹狀數組區間操作+單點操作。
假設數組a是原數組，b是a的差分數組，由前綴和的定義，我們求前n個元素的和，即

先對原數組a進行差分，得到差分數組b，纔對b[i]和b[i]*i分別維護一個樹狀數組tb和tc，
而l到r的和變爲
[(r + 1) * sum(tb, r) - sum(tc, r)]-[(l + 1) * sum(tb, l) - sum(tc, l)]
需要注意的是開vector<>後靠resize()維護大小會卡在60%

#pragma GCC optimize(2)
#include <bits/stdc++.h>
#define endl  '\n'
#define ll long long
using namespace std;
ll a[100005],b[100005];//a爲原數組，b爲a的差分數組
ll tb[100005],tc[100005]; //b[i]和b[i]*i分別維護的一個樹狀數組
int n,m; //n爲樹狀數組大小
int lowbit(int x)
{
    return x&-x;
}
void add(int i,ll x)
{
    ll v=i*x;
    while(i<=n)
    {
        tb[i]+=x;
        tc[i]+=v;
        i+=lowbit(i);
    }
}
void range_add(int l,int r,ll x)
{
    add(l,x);
    add(r+1,-x);
}
ll sum_t(ll arr[],int i)
{
    ll res=0;
    while(i>0)
    {
        res+=arr[i];
        i-=lowbit(i);
    }
    return res;
}
ll sum(int i)
{
    return (i+1)*sum_t(tb,i)-sum_t(tc,i);
}
ll range_sum(int l,int r)
{
    return sum(r)-sum(l-1);
}
 
int main() {
    ios::sync_with_stdio(0);cin.tie(0), cout.tie(0);
       cin>>n>>m;
       for(int i=1;i<=n;i++)
           cin>>a[i];
    for(int i=1;i<=n;i++)
           b[i]=a[i]-a[i-1];
    for(int i=1;i<=n;i++)
        add(i,b[i]);
    while(m--)
    {
        char c;cin>>c;
        if(c=='Q'){
            int x,y;cin>>x>>y;
            cout<<range_sum(x,y)<<endl;
        }
        else if(c=='C'){
            ll x,y,z;cin>>x>>y>>z;
            range_add(x,y,z);
        }
    }
    return 0;
}