poj-3468 A Simple Problem with Integers(线段树区间更新模板题）

Description

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

Sample Output

4
55
9
15

Hint

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

题意：长度为n的数组，C时在下标a,b内加c,Q时求小标a,b内数的和；

#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstring>
using namespace std;
const int MAX=1e5+5;
typedef long long ll;
ll a[MAX];
struct Node
{
    ll add;
    int l;
    int r;
    ll val;
}tree[4*MAX];
void pushUp(int root)
{
    tree[root].val=tree[root*2].val+tree[root*2+1].val;
}
void build(int root,int L,int R)
{
    tree[root].add=0;
    tree[root].l=L;
    tree[root].r=R;
    if(L==R)
    {
        tree[root].val=a[L];
        return;
    }
    int mid=(L+R)/2;
    build(root*2,L,mid);
    build(root*2+1,mid+1,R);
    pushUp(root);
}
void pushDown(int root)
{
    if(tree[root].add!=0)
    {
        tree[root*2].add+=tree[root].add;
        tree[root*2+1].add+=tree[root].add;
        tree[root*2].val+=tree[root].add*(tree[root*2].r-tree[root*2].l+1);
        tree[root*2+1].val+=tree[root].add*(tree[root*2+1].r-tree[root*2+1].l+1);
        tree[root].add=0;
    }
}
void updateSome(int root,int L,int R,int addNum)
{
    if(L<=tree[root].l&&R>=tree[root].r)
    {
        tree[root].add+=addNum;
        tree[root].val+=addNum*(tree[root].r-tree[root].l+1);
        return ;
    }
    pushDown(root);
    int mid=(tree[root].l+tree[root].r)/2;
    if(L<=mid)updateSome(root*2,L,R,addNum);
    if(R>mid)updateSome(root*2+1,L,R,addNum);
    pushUp(root);
    return ;
}
ll query(int root,int L,int R)
{
    if(L<=tree[root].l&&R>=tree[root].r)
    {
        return tree[root].val;
    }
    if(R<tree[root].l||L>tree[root].r)
        return 0;
    pushDown(root);
    int mid=(tree[root].l+tree[root].r)/2;
    ll ans=0;
    if(L<=mid)ans+=query(root*2,L,R);
    if(R>mid)ans+=query(root*2+1,L,R);
    return ans;
}
int main()
{
    int n,q;
    while(cin>>n>>q)
    {
        for(int i=1;i<=n;i++)
            cin>>a[i];
        build(1,1,n);
        while(q--)
        {
            char que;
            int x,y;
            cin>>que>>x>>y;
            if(que=='Q')
            {
               cout<<query(1,x,y)<<endl;
            }else
            {
                ll addNum;
                cin>>addNum;
                updateSome(1,x,y,addNum);
            }
        }
    }
    return 0;
}

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<map>
#include<set>
#include<queue>
#include<stack>
using namespace std;
typedef long long ll;
const int maxn=1e5+5;
ll n,q;
ll a[maxn];
struct Node
{
     ll sum;
     ll add;
     ll l;
     ll r;
     Node()
     {
         sum=0;
         add=0;
     }
}tree[4*maxn];
void _sum(ll root)
{
    tree[root].sum=tree[root*2].sum+tree[root*2+1].sum;
}
void _down(ll root)
{
    if(tree[root].add!=0)
    {
        tree[root*2].sum+=(tree[root*2].r-tree[root*2].l+1)*tree[root].add;
        tree[root*2].add+=tree[root].add;
        tree[root*2+1].sum+=(tree[root*2+1].r-tree[root*2+1].l+1)*tree[root].add;
        tree[root*2+1].add+=tree[root].add;
        tree[root].add=0;
    }
    return ;
}
void build(ll root,ll L,ll R)
{
    tree[root].l=L;
    tree[root].r=R;
    if(L==R)
    {
        tree[root].sum=a[L];
        return ;
    }
    ll mid=(L+R)/2;
    build(root*2,L,mid);
    build(root*2+1,mid+1,R);
    _sum(root);
}
void update(ll root,ll L,ll R,ll addNum)
{
    if(L<=tree[root].l&&R>=tree[root].r)
    {

        tree[root].sum+=(tree[root].r-tree[root].l+1)*addNum;
        tree[root].add+=addNum;
        return ;
    }else if(R<tree[root].l||L>tree[root].r)return ;
    else
    {
        _down(root);
        update(root*2,L,R,addNum);
        update(root*2+1,L,R,addNum);
        _sum(root);
    }
    return ;
}
ll query(ll root,ll L,ll R)
{
    if(L<=tree[root].l&&R>=tree[root].r)
    {
       return tree[root].sum;
    }else if(L>tree[root].r||R<tree[root].l)return 0;
    else
    {
        _down(root);
        ll temp=0;
        temp+=query(root*2,L,R);
        temp+=query(root*2+1,L,R);
        return temp;
    }
}
int main()
{

    while(~scanf("%lld%lld",&n,&q))
    {
        for(ll i=1;i<=n;i++)
        {
            scanf("%lld",&a[i]);
        }
        build(1,1,n);

        while(q--)
        {
            getchar();
            char op;
            ll x,y;
            scanf("%c %lld %lld",&op,&x,&y);
            if(op=='Q')printf("%lld\n",query(1,x,y));
            else
            {
                ll addNum;
                scanf("%lld",&addNum);
                update(1,x,y,addNum);
            }
        }
    }
    return 0;
}