poj3468 a simple problem with integers

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.

直接用bit做会超时， 因为这里的加操作是对一个区间里的数都加

大白书上有线段树 和 bit的做法

直接对着敲的

感觉有的地方还不是那么理解 为什么会想到要这么做呢

树状数组ac代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<limits>
#include<stack>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<limits>
#include<stack>

using namespace std;

const int DAT_SIZE = (1 << 18) - 1;
const int MAX_N = 100005;
const int MAX_Q = 100005;
int N, Q;
int A[MAX_N];
char T[MAX_Q];
int L[MAX_Q], R[MAX_Q], X[MAX_Q];
long long data[DAT_SIZE], datb[DAT_SIZE];

void add(int a, int b, int x, int k, int l, int r)
{
    if(a <= l && r <= b){
        data[k] += x;
    }
    else if(l < b && a <r){
        datb[k] += (min(b,r) - max(a, l)) * x;
        add(a, b, x, k * 2 + 1, l, (l + r) / 2);
        add(a, b, x, k * 2 + 2, (l + r) / 2, r);
    }
}

long long sum(int a, int b, int k, int l, int r)
{
    if(b <= l || r <= a){
        return 0;
    }
    else if (a <= l && r <= b){
        return data[k] * (r - l) + datb[k];
    }
    else{
        long long res = (min(b, r) - max(a, l)) * data[k];
        res += sum(a, b, k * 2 + 1, l, (l + r) / 2);
        res += sum(a, b, k * 2 + 2, (l + r) / 2, r);
        return res;
    }
}

void solve()
{
    for(int i = 0; i < N; i++){
        add(i, i + 1, A[i], 0, 0, N);
    }
    for(int i = 0; i < Q; i++){
        if(T[i] == 'C'){
            add(L[i], R[i] + 1, X[i], 0, 0, N);
        }
        else{
            printf("%lld\n",sum(L[i], R[i] + 1, 0, 0, N));
        }
    }
}

int main()
{
    while(scanf("%d%d",&N,&Q) != EOF){
        for(int i = 0; i < N; i++){
            scanf("%d",&A[i]);
        }
        for(int i = 0; i < Q; i++){
            getchar();
            scanf("%c",&T[i]);
            if(T[i] == 'C'){
                scanf("%d%d%d",&L[i],&R[i],&X[i]);
            }
            else{
                scanf("%d%d",&L[i],&R[i]);
            }
        }
        solve();
    }
    return 0;
}