POJ3468_A Simple Problem with Integers_树状数组区间和::区间修改

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A Simple Problem with Integers

Time Limit: 5000MS Memory Limit: 131072KTotal Submissions: 109152 Accepted: 33998Case 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 51 2 3 4 5 6 7 8 9 10Q 4 4Q 1 10Q 2 4C 3 6 3Q 2 4

Sample Output

Hint

The sums may exceed the range of 32-bit integers.

给一个含有 N 个元素的数组，再给 M 个操作。其中 M 个操作为求区间和或者区间元素统一加上一个数。

对于区间和询问操作，用树状数组维护每个元素的前缀和，用前缀和之差来表示区间和。

对于区间修改操作，有

i<l : s'(i) = s(i)

l <= i <= r : s'(i) = s(i) + x*(i-l-1) = s(i) + x*i - x*(l-1)

I>r : s'(i) = s(i) + x*(r-l-1) = s(i) + x*r - x*(l-1)

这时对区间的修改只需要对区间两端点操作就可以了，需要维护两个树状数组：

一个保存原数组，区间首的 -x(l-1) 和区间尾的 x*r，这两项是与 i 无关的常数项

一个保存区间首的 x 和区间尾的 -x，这两项是与 i 有关的

具体分析见《挑战程序设计竞赛》181页。

树状数组的基本原理见《挑战程序设计竞赛》174页。

#include<cstdio>#include<iostream>using namespace std;typedef long long LL;const int maxn = 100000 + 10;LL Bit[maxn];//第一个树状数组LL Bit1[maxn];//第二个树状数组int N, M;char C;//单点更改void Add(LL *B, int p, int x){while(p <= N){B[p] += x;p += p&-p;}}//求前缀和LL Sum(LL *B, int p){LL res = 0;while(p){res += B[p];p -= p&-p;}return res;}int main (){scanf("%d %d", &N, &M);for(int i= 1; i<= N; i++){LL temp;scanf("%lld", &temp);Add(Bit, i, temp);}for(int k= 0; k< M; k++){scanf(" %c", &C);if(C == 'C'){int l, r;LL x;scanf("%d %d %lld", &l, &r, &x);//对区间首尾的四项操作Add(Bit, l, -(l-1)*x);Add(Bit1, l, x);Add(Bit, r+1, x*r);Add(Bit1, r+1, -x);}else{int l, r;scanf("%d %d", &l, &r);//分别求前缀和再作差LL res = 0;res += Sum(Bit, r) + Sum(Bit1, r) * r;res -= Sum(Bit, l-1) + Sum(Bit1, l-1) * (l-1);printf("%lld\n", res);}}return 0;}

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