POJ 3468 A Simple Problem with Integers（线段树）

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 126311		Accepted: 39238
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

#include <iostream>
#include <cstring>
#include <string>
#include <algorithm>
typedef long long ll;
using namespace std;
int n, q;
struct node
{
    int l, r;
    ll date;//区间内每个数都要加的数
    ll datb;//数据的和
}a[400005];


void build(int l,int r,int k)
{
    a[k].l = l; a[k].r = r;
    if (r == l)
    {
        scanf("%lld", &a[k].datb);
        return;
    }
    int mid = (l + r) / 2;//传的是l和r，不是a[k].l和a[k].r
    build(l, mid , 2 * k);
    build(mid + 1, r, 2 * k + 1);
    a[k].datb = a[2 * k].datb + a[2 * k + 1].datb;
}

void add(int k,int l,int r,int v)
{
    if (a[k].l >= l && r >= a[k].r)//如果区间刚好在a[k].l和a[k].r之间
        a[k].date += v;//每个数都要加，所以放到date中
    else if(a[k].l<=r&&l<=a[k].r)//部分重叠的话
    {
        a[k].datb+= (min(r, a[k].r) - max(l, a[k].l)+1)*v;//我加号居然忘了
        add(2 * k, l,r, v);
        add(2 * k + 1,l, r, v);
    }
    //否则跳出循环
}

ll Sum(int l, int r,int k)
{
    if (l > a[k].r||r < a[k].l)
        return 0;
    else if (l <= a[k].l&&r >= a[k].r)
    {
        /*cout << "a[k].l " << a[k].l << " a[k].r " << a[k].r << endl;
        cout << a[k].datb + a[k].date *(a[k].r - a[k].l + 1) << endl;*/
        return a[k].datb + a[k].date *(a[k].r - a[k].l + 1);
    }
    else
    {
        ll ans = (min(a[k].r, r) - max(a[k].l, l)+1)*a[k].date;
        ans += Sum(l, r, k * 2);//传的是l和r，不是a[k].l和a[k].r
        ans += Sum(l, r, k * 2 + 1);
        return ans;
    }
}

int main()
{
    while (~scanf("%d%d", &n, &q))
    {
        for (int i = 0; i <= 400000; i++)
            a[i].datb = a[i].date = 0;
        build(1, n, 1);
        while (q--)
        {
            char s;
            int x, y;
            cin >> s >> x >> y;
            if (s == 'C')
            {
                int z;
                cin >> z;
                add(1, x, y, z);
            }
            else
            {
                ll sum = Sum(x, y, 1);
                printf("%lld
", sum);
            }
        }
    }
    return 0;
}