BZOJ-3212 Pku3468 A Simple Problem with Integers 裸线段树区间维护查询

3212: Pku3468 A Simple Problem with Integers
Time Limit: 1 Sec Memory Limit: 128 MB
Submit: 1278 Solved: 560
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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.

题目大意：区间修改和区间查询

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代码如下：

/**************************************************************
    Problem: 3212
    User: DaD3zZ
    Language: C++
    Result: Accepted
    Time:48 ms
    Memory:8304 kb
****************************************************************/

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define maxn 100001
long long sum[maxn<<2]={0},delta[maxn<<2]={0};
long long a[maxn]={0};

void updata(int now)
{
    sum[now]=sum[now<<1]+sum[now<<1|1];
}

void build(int now,int l,int r)
{
    if (l==r)
        {
            sum[now]=a[l];
            return;
        }
    int mid=(l+r)>>1;
    build(now<<1,l,mid);
    build(now<<1|1,mid+1,r);
    updata(now);
}

void pushdown(int now,int ln,int rn)
{
    if (delta[now]!=0)
        {
            delta[now<<1]+=delta[now];
            delta[now<<1|1]+=delta[now];
            sum[now<<1]+=delta[now]*ln;
            sum[now<<1|1]+=delta[now]*rn;
            delta[now]=0;
        }
}

void change(int L,int R,int now,int l,int r,long long data)
{
    if (L<=l && R>=r)
        {
            sum[now]+=data*(r-l+1);
            delta[now]+=data;
            return;
        }
    int mid=(l+r)>>1;
    pushdown(now,mid-l+1,r-mid);//这里需要先下放一波标记不然会出错
    if (L<=mid)
        change(L,R,now<<1,l,mid,data);
    if (R>mid)
        change(L,R,now<<1|1,mid+1,r,data);
    updata(now);
}

long long query(int L,int R,int now,int l,int r)
{
    if (L<=l && R>=r)
        return sum[now];
    int mid=(l+r)>>1;
    pushdown(now,mid-l+1,r-mid);
    long long total=0;
    if (L<=mid)
        total+=query(L,R,now<<1,l,mid);
    if (R>mid)
        total+=query(L,R,now<<1|1,mid+1,r);
    return total;
}

int main()
{
    int n,m;
    scanf("%d%d",&n,&m);
    for (int i=1; i<=n; i++)
        scanf("%lld",&a[i]);
    build(1,1,n);
    for (int i=1; i<=m; i++)
        {
            char c[3];
            int l,r;
            scanf("%s",&c);
            scanf("%d%d",&l,&r);
            if (c[0]=='Q')
                {
                    long long ans=query(l,r,1,1,n);
                    printf("%lld\n",ans);
                }
            else
                {
                    long long data;
                    scanf("%lld",&data);
                    change(l,r,1,1,n,data);
                }
        }
    return 0;   
}