这个题其实有O(n)的算法，不过还是用线段树做了一遍，还写了个自认为不错的pushalldown函数，哈哈。
1 #include <iostream>
2 #include <cstring>
3 #include <cstdio>
4 using namespace std;
5
6 const int N = 100001;
7 int ans[N];
8
9 struct Node
10 {
11 int l, r, add;
12 } node[N << 2];
13
14 void build( int i, int l, int r )
15 {
16 node[i].l = l, node[i].r = r, node[i].add = 0;
17 if ( l == r ) return ;
18 int lc = i << 1, rc = lc | 1, mid = ( l + r ) >> 1;
19 build( lc, l, mid );
20 build( rc, mid + 1, r );
21 }
22
23 void pushdown( int i )
24 {
25 if ( node[i].add )
26 {
27 int lc = i << 1, rc = lc | 1;
28 node[lc].add += node[i].add;
29 node[rc].add += node[i].add;
30 node[i].add = 0;
31 }
32 }
33
34 void update( int i, int l, int r )
35 {
36 if ( node[i].l == l && node[i].r == r )
37 {
38 node[i].add++;
39 return ;
40 }
41 //可以最后一起放下来
42 //pushdown(i);
43 int lc = i << 1, rc = lc | 1, mid = ( node[i].l + node[i].r ) >> 1;
44 if ( r <= mid )
45 {
46 update( lc, l, r );
47 }
48 else if ( l > mid )
49 {
50 update( rc, l, r );
51 }
52 else
53 {
54 update( lc, l, mid );
55 update( rc, mid + 1, r );
56 }
57 }
58
59 void pushalldown( int i )
60 {
61 if ( node[i].l == node[i].r )
62 {
63 ans[node[i].l] = node[i].add;
64 return ;
65 }
66 pushdown(i);
67 pushalldown( i << 1 );
68 pushalldown( i << 1 | 1 );
69 }
70
71 int main ()
72 {
73 int n;
74 while ( scanf("%d", &n), n )
75 {
76 build( 1, 1, n );
77 for ( int i = 1; i <= n; i++ )
78 {
79 int a, b;
80 scanf("%d%d", &a, &b);
81 update( 1, a, b );
82 }
83 pushalldown(1);
84 for ( int i = 1; i <= n; i++ )
85 {
86 printf("%d", ans[i]);
87 if ( i != n ) putchar(' ');
88 else putchar('\n');
89 }
90 }
91 return 0;
92 }
来源：https://www.cnblogs.com/huoxiayu/p/4690931.html