POJ 3468 树状数组

题目

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

题意

给一个数组，可以更改任一个元素的值。问任意一个区间内的元素和

题解

用线段树当然是可以做的。这里用树状数组，树状数组实质上是用二进制的存储方法求前缀和。然后查找和更改都是O（logn）。

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <stack>#include <string>#include <set>#include <cmath>#include <map>#include <queue>#include <sstream>#include <vector>#define m0(a) memset(a,0,sizeof(a))#define mm(a) memset(a,0x3f,sizeof(a))#define m_1(a) memset(a,-1,sizeof(a))#define f(i,a,b) for(i = a;i<=b;i++)#define fi(i,a,b) for(i = a;i>=b;i--)#define FFR freopen("data.in","r",stdin)#define FFW freopen("data.out","w",stdout)#define lowbit(k) ((k)&(-k))using namespace std;#define SIZE 100000typedef long long ll;ll a[SIZE+10];ll bit1[SIZE+10];ll bit2[SIZE+10];void init(int n){    int i,j;    a[0] = 0;    f(i,1,n)        a[i] += a[i-1];}ll bit_sum(ll* bit,int n){    ll sum = 0;    while(n){        sum+=bit[n];        n-=lowbit(n);    }    return sum;}void bit_add(ll* bit,int n,int x,int k){    while(x<=n){        bit[x]+=k;        x+=lowbit(x);    }}int main(){    ios_base::sync_with_stdio(false); cin.tie(0);    int n,m;    cin>>n>>m;    int i,j;    f(i,1,n) cin>>a[i];    init(n);    char c;    f(i,1,m){        cin>>c;        if(c=='Q'){            int L,R;            cin>>L>>R;            ll ans = a[R]-a[L-1];            ans+=(R+1)*bit_sum(bit1,R)-bit_sum(bit2,R);            ans-=L*bit_sum(bit1,L-1)-bit_sum(bit2,L-1);            cout<<ans<<endl;        }else{            int k,L,R;            cin>>L>>R>>k;            bit_add(bit1,n,L,k);bit_add(bit1,n,R+1,-k);            bit_add(bit2,n,L,k*L);bit_add(bit2,n,R+1,-k*(R+1));        }    }    return 0;}