A Simple Problem with Integers（poj 3468）

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 130621		Accepted: 40554
Case Time Limit: 2000MS

Description

You have N integers, A₁, A₂, ... , A_N. 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, ... , A_N. -1000000000 ≤ A_i ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of A_a, A_a+1, ... , A_b. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of A_a, A_a+1, ... , A_b.

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.

Source

POJ Monthly--2007.11.25, Yang Yi

转发自：https://blog.csdn.net/Acceptedxukai/article/details/6933446

#include<cstdio>
#include<algorithm>
using namespace std;
const int N=1e5+5;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
long long sum[N<<2],add[N<<2];
struct Node
{
    int l,r;
    int mid()
    {
        return (l+r)>>1;
    }
}tree[N<<2];
void PushUp(int rt)//通过当前节点rt把值递归向上更新到根节点
{
    sum[rt]=sum[rt<<1]+sum[rt<<1|1];
}
void PushDown(int rt,int m)//通过当前节点rt递归向下去更新rt子节点的值
{
    if(add[rt])
    {
        add[rt<<1]+=add[rt];
        add[rt<<1|1]+=add[rt];
        sum[rt<<1]+=add[rt]*(m-(m>>1));
        sum[rt<<1|1]+=add[rt]*(m>>1);
        add[rt]=0;
    }
}
void build(int l,int r,int rt)
{
    tree[rt].l=l;
    tree[rt].r=r;
    add[rt]=0;
    if(l==r)
    {
        scanf("%lld",&sum[rt]);
        return;
    }
    int m=tree[rt].mid();
    build(lson);
    build(rson);
    PushUp(rt);
}
void update(int c,int l,int r,int rt)
{
    if(tree[rt].l==l&&tree[rt].r==r)
    {
        add[rt]+=c;
        sum[rt]+=(long long)c*(r-l+1);
        return;
    }
    if(tree[rt].l==tree[rt].r) return;
    PushDown(rt,tree[rt].r-tree[rt].l+1);
    int m=tree[rt].mid();
    if(r<=m) update(c,l,r,rt<<1);
    else if(l>m) update(c,l,r,rt<<1|1);
    else
    {
        update(c,l,m,rt<<1);
        update(c,m+1,r,rt<<1|1);
    }
    PushUp(rt);
}
long long query(int l,int r,int rt)
{
    if(l==tree[rt].l&&r==tree[rt].r) return sum[rt];
    PushDown(rt,tree[rt].r-tree[rt].l+1);
    int m=tree[rt].mid();
    long long res=0;
    if(r<=m) res+=query(l,r,rt<<1);
    else if(l>m) res+=query(l,r,rt<<1|1);
    else
    {
        res+=query(l,m,rt<<1);
        res+=query(m+1,r,rt<<1|1);
    }
    return res;
}
int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)==2)
    {
        build(1,n,1);
        while(m--)
        {
            char ch[2];
            scanf("%s",ch);
            int a,b,c;
            if(ch[0]=='Q')
            {
                scanf("%d%d",&a,&b);
                printf("%lld\n",query(a,b,1));
            }
            else
            {
                scanf("%d%d%d",&a,&b,&c);
                update(c,a,b,1);
            }
        }
    }
    return 0;
}