筱玛爱线段树

题目地址：

https://ac.nowcoder.com/acm/problem/25737

基本思路：

参考代码：

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>
using namespace std;
#define IO std::ios::sync_with_stdio(false); cin.tie(0)
#define int long long
#define ull unsigned long long
#define SZ(x) ((int)(x).size())
#define all(x) (x).begin(), (x).end()
#define rep(i, l, r) for (int i = l; i <= r; i++)
#define per(i, l, r) for (int i = l; i >= r; i--)
#define mset(s, _) memset(s, _, sizeof(s))
#define pb push_back
#define pii pair <int, int>
#define mp(a, b) make_pair(a, b)
#define INF 0x3f3f3f3f

inline int read() {
  int x = 0, neg = 1; char op = getchar();
  while (!isdigit(op)) { if (op == '-') neg = -1; op = getchar(); }
  while (isdigit(op)) { x = 10 * x + op - '0'; op = getchar(); }
  return neg * x;
}
inline void print(int x) {
  if (x < 0) { putchar('-'); x = -x; }
  if (x >= 10) print(x / 10);
  putchar(x % 10 + '0');
}

const int maxn = 1e5 + 10;
const int mod = 1e9 + 7;
int n,m,op[maxn],l[maxn],r[maxn],ans[maxn];
void add(int &x,int v) { x += v; if (x >= mod) x -= mod; }
struct BIT{
    int t[maxn << 2];
    int lowbit(int x){ return x & (-x); }
    void clear(){ memset(t,0,sizeof(t)); }
    int sum(int x){ int res = 0; while(x){ add(res,t[x]); x -= lowbit(x); } return res; }
    void update(int x,int v){ while(x <= m){ add(t[x],v); x += lowbit(x); } }
}bit;
signed main() {
  IO;
  cin >> n >> m;
  rep(i, 1, m) cin >> op[i] >> l[i] >> r[i];
  for (int i = m; i >= 1; i--) {
    int val = (bit.sum(i) + 1) % mod; // 操作的次数,注意开始本身就要进行一次操作;
    if (op[i] == 2) bit.update(l[i], val), bit.update(r[i] + 1, mod - val); // 差分维护区间的操作数;
    else if (op[i] == 1) add(ans[l[i]], val), add(ans[r[i] + 1], mod - val); // 差分维护结果数组;
  }
  for (int i = 1; i <= n; i++) {
    add(ans[i],ans[i-1]);
    cout << ans[i] << ' ';
  }
  cout << '\n';
  return 0;
}