3468 A Simple Problem with Integers

 

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Time Limit: 5000MS　　Memory Limit: 131072K　　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.

 

 

这是一道树状数组-[区间更新 区间查询]模板题，详情点击这里.

以下，A[i]代表原数组，D[j]代表差分数组.

 ∑ⁿ_{i = 1}A[i] = ∑ⁿ_{i = 1} ∑ⁱ_{j = 1}D[j];

 则 A[1]+A[2]+...+A[n]

   = (D[1]) + (D[1]+D[2]) + ... + (D[1]+D[2]+...+D[n]) 

   = n*D[1] + (n-1)*D[2] +... +D[n]

   = n * (D[1]+D[2]+...+D[n]) - (0*D[1]+1*D[2]+...+(n-1)*D[n])

所以上式可以变为∑ⁿ_{i = 1}A[i] = n*∑ⁿ_{i = 1}D[i] - ∑ⁿ_{i = 1}( D[i]*(i-1) );

于是，维护两个树状数组，sum1[i] = D[i]，sum2[i] = D[i]*(i-1);

※注意：题目中也有提醒道，本题数据较大，要用long long.

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <cmath>
#include <algorithm>
#define INF 0x3f3f3f3f
#define zero 1e-7

using namespace std;
typedef long long ll;
const ll mod=1000000007;
const ll max_n=100005;
ll a[max_n]={0};
ll sum1[max_n]={0}, sum2[max_n]={0};
ll n;

ll lowbit(ll x) {
    return x&(-x);
}

void update(ll i, ll k) {
    ll temp=i;
    while(i<=n) {
        sum1[i]+=k;
        sum2[i]+=(temp-1)*k;
        i+=lowbit(i);
    }
}

ll getsum(ll i) {
    ll res=0;
    ll temp=i;
    while(i>0) {
        res+=temp*sum1[i]-sum2[i];
        i-=lowbit(i);
    }
    return res;
}

int main() {
    ll q;
    cin>>n>>q;
    for(ll i=1; i<=n; i++) {
        scanf("%I64d", &a[i]);
        update(i, a[i]-a[i-1]);
    }
    char opt;
    ll a, b, c;
    for(ll i=0; i<q; i++) {
        scanf(" %c", &opt);
        if(opt=='C') {
            scanf("%I64d %I64d %I64d", &a, &b, &c);
            update(a, c);
            update(b+1, -c);
        }
        else {
            scanf("%I64d %I64d", &a, &b);
            printf("%I64d\n", getsum(b)-getsum(a-1));
        }
    }
    return 0;
}