Problem:
Range Sum Query

Write a program which manipulates a sequence A = {a1, a2, . . . , an} with the following operations:

add(i, x): add x to ai.

getSum(s, t): print the sum of as, as+1,...,at.

Note that the initial values of ai (i = 1, 2, . . . , n) are 0.

Input

n q
com1 x1 y1
com2 x2 y2
...
comq xq yq

In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, q queries are given where com represents the type of queries. '0' denotes add(xi, yi) and '1' denotes getSum(xi, yi).

Output

For each getSum operation, print the sum in a line.

Constraints

1 ≤ n ≤ 100000

1 ≤ q ≤ 100000

If comi is 0, then 1 ≤ xi ≤ n, 0 ≤ yi ≤ 1000.

If comi is 1, then 1 ≤ xi ≤ n, 1 ≤ yi ≤ n.

Sample Input 1

3 5
0 1 1
0 2 2
0 3 3
1 1 2
1 2 2

Sample Output 1

3
2

Solution in C++14 (judge verdict: Accepted)

#include <bits/stdc++.h>

//{ START
using namespace std;
#define int int64_t
#define rep(i, a, n) for (int i = (a); i < (n); ++i)
#define reps(i, a, n) for (int i = (a); i > (n); --i)
#define arep(i, x) for (auto &&i : (x))
#define irep(i, x) for (auto i = (x).begin(); i != (x).end(); ++i)
//降順はgreater<T>()
#define all(x) (x).begin(), (x).end()
#define rv(s) reverse((s).begin(), (s).end())
// gcd lcmはそのままok
#define gcd(a, b) __gcd(a, b)
#define bits(n) (1LL << (n))
#define pcnt(x) __builtin_popcountll(x)
//配列内等要素削除
#define unique(x) (x).erase(unique((x).begin(), (x).end()), (x).end())
#define Fixed(n) fixed << setprecision(n)
//総和
#define sowa(n) ((n) * ((n) + 1)) / 2
#define pb(x, a) (x).push_back(a)
#define fb(x, a) (x).push_front(a)
#define eb(x, a) (x).emplace_back(a)
#define F first
#define S second
#define cauto const auto &
using P = pair<int, int>;
using Graph = vector<vector<P>>;
template <class T>  //昇順
using min_heap = priority_queue<T, vector<T>, greater<T>>;
template <class T>  //降順
using max_heap = priority_queue<T>;
template <class A, class B>
using umap = unordered_map<A, B>;
template <class A>
using uset = unordered_set<A>;
template <typename A, size_t N, typename T>
void Fill(A (&array)[N], const T &val) {  //多次元初期化
  std::fill((T *)array, (T *)(array + N), val);
}
template <class T>
bool chmax(T &a, const T &b) {  //最大値更新 返り値はbool
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool chmin(T &a, const T &b) {  //最小値更新 返り値はbool
  if (b < a) {
    a = b;
    return 1;
  }
  return 0;
}
int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};
constexpr int INF = 0x3f3f3f3f;
constexpr int LINF = 0x3f3f3f3f3f3f3f3fLL;
constexpr int mod1 = 1e9 + 7;
constexpr int mod2 = 998244353;
//} END

/* BIT: 区間和の更新や計算を行う構造体
初期値は a_1 = a_2 = ... = a_n = 0
add(i,x): a_i += x とする
sum(i): a_1 + a_2 + ... + a_i を計算する
計算量は全て O(logn)
*/
template <typename T>
struct BIT {
  int n;          // 配列の要素数(数列の要素数+1)
  vector<T> bit;  // データの格納先
  BIT(int n_) : n(n_ + 1), bit(n, 0) {}

  void add(int i, T x) {
    for (int idx = i; idx < n; idx += (idx & -idx)) bit[idx] += x;
  }

  T sum(int i) {
    T s(0);
    for (int idx = i; idx > 0; idx -= (idx & -idx)) s += bit[idx];
    return s;
  }
};

signed main(){
  cin.tie(nullptr);
  ios_base::sync_with_stdio(false);

  int n,q;
  cin>>n>>q;
  BIT<int> bit(n);

  rep(i,0,q){
    int c,x,y;
    cin>>c>>x>>y;
    if(c==1){
      cout<<bit.sum(y)-bit.sum(x-1)<<'\n';
    }else{
      bit.add(x,y);
    }
  }

  return 0;
}