“反转类问题”需要注意的是pushdown懒标记的时候要用异或（或者++%2）而不是赋值，因为它是和奇偶性有关系的。
1 #include <iostream> 2 #include <cstring> 3 #include <cstdio> 4 using namespace std; 5 6 const int N = 100001; 7 8 struct Node 9 { 10 int l, r, sum, rev; 11 } node[N << 2]; 12 13 void pushup( int i ) 14 { 15 node[i].sum = node[i << 1].sum + node[i << 1 | 1].sum; 16 } 17 18 void build( int i, int l, int r ) 19 { 20 node[i].l = l, node[i].r = r, node[i].rev = node[i].sum = 0; 21 if ( l == r ) return ; 22 int mid = ( l + r ) >> 1; 23 build( i << 1, l, mid ); 24 build( i << 1 | 1, mid + 1, r ); 25 } 26 27 void pushdown( int i ) 28 { 29 if ( node[i].rev ) 30 { 31 int lc = i << 1, rc = lc | 1; 32 node[lc].rev ^= 1; 33 node[rc].rev ^= 1; 34 node[lc].sum = node[lc].r - node[lc].l + 1 - node[lc].sum; 35 node[rc].sum = node[rc].r - node[rc].l + 1 - node[rc].sum; 36 node[i].rev = 0; 37 } 38 } 39 40 void update( int i, int l, int r ) 41 { 42 if ( node[i].l == l && node[i].r == r ) 43 { 44 node[i].sum = node[i].r - node[i].l + 1 - node[i].sum; 45 node[i].rev ^= 1; 46 return ; 47 } 48 pushdown(i); 49 int mid = ( node[i].l + node[i].r ) >> 1; 50 if ( r <= mid ) 51 { 52 update( i << 1, l, r ); 53 } 54 else if ( l > mid ) 55 { 56 update( i << 1 | 1, l, r ); 57 } 58 else 59 { 60 update( i << 1, l, mid ); 61 update( i << 1 | 1, mid + 1, r ); 62 } 63 pushup(i); 64 } 65 66 int query( int i, int l, int r ) 67 { 68 if ( node[i].l == l && node[i].r == r ) 69 { 70 return node[i].sum; 71 } 72 pushdown(i); 73 int mid = ( node[i].l + node[i].r ) >> 1; 74 if ( r <= mid ) 75 { 76 return query( i << 1, l, r ); 77 } 78 else if ( l > mid ) 79 { 80 return query( i << 1 | 1, l, r ); 81 } 82 else 83 { 84 return query( i << 1, l, mid ) + query( i << 1 | 1, mid + 1, r ); 85 } 86 } 87 88 int main () 89 { 90 int n, m; 91 while ( scanf("%d%d", &n, &m) != EOF ) 92 { 93 build( 1, 1, n ); 94 while ( m-- ) 95 { 96 int op, x, y; 97 scanf("%d%d%d", &op, &x, &y); 98 if ( op == 0 ) 99 { 100 update( 1, x, y ); 101 } 102 else 103 { 104 printf("%d ", query( 1, x, y )); 105 } 106 } 107 } 108 return 0; 109 }