poj 3468 A Simple Problem with Integers

http://poj.org/problem?id=3468

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 75149		Accepted: 23149
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.

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<algorithm>
#define N 100010
#define Lson root<<1, L, tree[root].Mid()
#define Rson root<<1|1, tree[root].Mid() + 1, R

using namespace std;

struct Tree
{
    int L, R;
    long long sum, e;
    bool op;
    int Mid()
    {
        return (L + R) / 2;
    }
    int Len()
    {
        return (R - L + 1);
    }
} tree[N * 4];

long long al[N];

void Update(int root)
{
    if(tree[root].op && tree[root].L != tree[root].R)
    {
        tree[root].op = false;
        tree[root<<1].op = tree[root<<1|1].op = true;

        tree[root<<1].e += tree[root].e;
        tree[root<<1|1].e += tree[root].e;

        tree[root<<1].sum += tree[root<<1].Len() * tree[root].e;
        tree[root<<1|1].sum += tree[root<<1|1].Len() * tree[root].e;

        tree[root].e = 0;
    }
}

void Build(int root, int L, int R)
{
    tree[root].L = L, tree[root].R = R;
    tree[root].op = false;
    if(L == R)
    {
        tree[root].sum = al[L];
        return ;
    }
    Build(Lson);
    Build(Rson);

    tree[root].sum = tree[root<<1].sum + tree[root<<1|1].sum;
}

void Insert(int root, int L, int R, long long e)
{
    Update(root);
    tree[root].sum += (R - L + 1) * e;
    if(tree[root].L == L && tree[root].R == R)
    {
        tree[root].op = true;
        tree[root].e = e;
        return ;
    }
    if(R <= tree[root].Mid())
        Insert(root<<1, L, R, e);
    else if(L > tree[root].Mid())
        Insert(root<<1|1, L, R, e);
    else
    {
        Insert(Lson, e);
        Insert(Rson, e);
    }
}

long long Query(int root, int L, int R)
{
    Update(root);
    if(tree[root].L == L && tree[root].R == R)
        return tree[root].sum;
    if(R <= tree[root].Mid())
        return Query(root<<1, L, R);
    else if(L > tree[root].Mid())
        return Query(root<<1|1, L, R);
    else
        return Query(Lson) + Query(Rson);
}

int main()
{
    int n, q, a, b, i;
    long long c;
    char s[10];
    while(scanf("%d%d", &n, &q) != EOF)
    {
        for(i = 1 ; i <= n ; i++)
            scanf("%I64d", &al[i]);
        Build(1, 1, n);
        while(q--)
        {
            scanf("%s%d%d", s, &a, &b);
            if(s[0] == 'Q')
                printf("%I64d\n", Query(1, a, b));
            else
            {
                scanf("%I64d", &c);
                Insert(1, a, b, c);
            }
        }
    }
    return 0;
}