C - A Simple Problem with Integers

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.

这题应该算是线段树区间修改的模板题了，一开始因为longlong没开wa了几发，ac了后又打了两遍，毕竟刚学线段树，增加熟练度

#include<iostream>
#include<stdio.h>

using namespace std;

const int MAXN = 100000 + 10;
struct f
{
    long long len ;
    long long val;
    long long lazy;
}a[MAXN * 4];
int n , m ;

void pushup(int k)
{
    a[k].val = a[k << 1].val + a[k << 1 | 1].val;
}

void pushdown(int x)
{
    if(a[x].lazy)
    {
        a[x << 1].lazy += a[x].lazy;
        a[x << 1 | 1].lazy += a[x].lazy;
        a[x << 1].val += a[x << 1].len * a[x].lazy;
        a[x << 1 | 1].val += a[x << 1 | 1].len * a[x].lazy;
        a[x].lazy = 0;
    }
}

void build(int left , int right , int rt)
{
    a[rt].len = right - left + 1;
    if(left == right)
    {
        scanf("%lld" , &a[rt].val);
        return ;
    }
    int mid = (left + right) >> 1;
    build(left , mid , rt << 1);
    build(mid + 1 , right , rt << 1 | 1);
    a[rt].val = a[rt << 1].val + a[rt << 1 | 1].val;
}

long long que(int x , int y ,int left , int right ,int rt)
{
    if(left >= x && right <= y)
    {
        return a[rt].val;
    }
    if(a[rt].lazy)
    {
        pushdown(rt);
    }
    long long sum = 0;
    int mid = (left + right) >> 1;
    if(mid >= x)
        sum += que(x , y , left , mid , rt << 1);
    if(mid < y)
        sum += que(x , y , mid + 1, right , rt << 1 | 1);
    return sum;
}

void update(int x , int y ,int k , int left , int right ,int rt)
{
    if(left >= x && right <= y)
    {
        a[rt].lazy += k;
        a[rt].val += a[rt].len * k;
        return;
    }
    if(a[rt].lazy)
        pushdown(rt);
    int mid = (left + right) >> 1;
    if(mid >= x)
        update(x , y , k , left , mid , rt << 1);
    if(mid < y)
        update(x , y , k , mid + 1, right , rt << 1 | 1);
    a[rt].val = a[rt << 1].val + a[rt << 1 | 1].val;
}

int main()
{
    scanf("%d %d" , &n , &m);
    build(1 , n , 1);
    char q;
    int x , y , k;
    while(m --)
    {
        cin >> q;
        if(q == 'Q')
        {
            scanf("%d %d" , &x , &y);
            printf("%lld\n" , que(x , y , 1, n , 1));
        }
        else
        {
            scanf("%d %d %d", &x , &y , &k);
            update(x , y , k , 1 , n , 1);
        }
    }
    return 0;
}