HDU 4351 Digital root 线段树区间合并

依然不是十分理解……待考虑……

 

  1 #include <cstdio>

  2 #include <cstring>

  3 #include <cstdlib>

  4 #include <algorithm>

  5 

  6 #define lson l, m, rt << 1

  7 #define rson m + 1, r, rt << 1 | 1

  8 

  9 using namespace std;

 10 

 11 const int MAXN = 122222;

 12 

 13 struct node

 14 {

 15     int num;

 16     int lnode, rnode;

 17     int root;

 18     node(): num(0), lnode(0), rnode(0), root(0){ }

 19 };

 20 

 21 int root[ MAXN << 2 ];

 22 int num[ MAXN << 2 ];

 23 int lnode[ MAXN << 2 ];

 24 int rnode[ MAXN << 2 ];

 25 int all[ 1 << 11 ][ 1 << 11 ];

 26 

 27 int GetRoot( int a )

 28 {

 29     if ( !a ) return 0;

 30     if ( a % 9 ) return a % 9;

 31     return 9;

 32 }

 33 

 34 void init()

 35 {

 36     for ( int i = 0; i < ( 1 << 10 ); ++i )

 37         for ( int j = 0; j < ( 1 << 10 ); ++j )

 38         {

 39             all[i][j] = 0;

 40             for ( int x = 0; x < 10; ++x )

 41                 if ( i & ( 1 << x ) )

 42                 {

 43                     for ( int y = 0; y < 10; ++y )

 44                         if ( j & ( 1 << y ) )

 45                             all[i][j] |= ( 1 << GetRoot( x + y ) );

 46                 }

 47         }

 48     return;

 49 }

 50 

 51 void PushUp( int rt )

 52 {

 53     int lc = rt << 1;

 54     int rc = rt << 1 | 1;

 55     root[rt] = root[lc] | root[rc] | all[ rnode[lc] ][ lnode[rc] ];

 56     num[rt] = all[ num[lc] ][ num[rc] ];

 57     lnode[rt] = lnode[lc] | all[ num[lc] ][ lnode[rc] ];

 58     rnode[rt] = rnode[rc] | all[ num[rc] ][ rnode[lc] ];

 59     return;

 60 }

 61 

 62 void build( int l, int r, int rt )

 63 {

 64     if ( l == r )

 65     {

 66         int a;

 67         scanf( "%d", &a );

 68         a = GetRoot(a);

 69         lnode[rt] = rnode[rt] = num[rt] = root[rt] = ( 1 << a );

 70         return;

 71     }

 72 

 73     int m = ( l + r ) >> 1;

 74     build( lson );

 75     build( rson );

 76     PushUp( rt );

 77     return;

 78 }

 79 

 80 node Query( int L, int R, int l, int r, int rt )

 81 {

 82     if ( L <= l && r <= R )

 83     {

 84         node D;

 85         D.root = root[rt];

 86         D.lnode = lnode[rt];

 87         D.rnode = rnode[rt];

 88         D.num = num[rt];

 89         return D;

 90     }

 91 

 92     int m = ( l + r ) >> 1;

 93 

 94     if ( R <= m ) return Query( L, R, lson );

 95     else if ( L > m ) return Query( L, R, rson );

 96     else

 97     {

 98         node LD, RD, D;

 99         LD = Query( L, R, lson );

100         RD = Query( L, R, rson );

101         D.root = LD.root | RD.root | all[ LD.rnode ][ RD.lnode ];

102         D.num = all[ LD.num ][ RD.num ];

103         D.lnode = LD.lnode | all[ LD.num ][ RD.lnode ];

104         D.rnode = RD.rnode | all[ RD.num ][ LD.rnode ];

105         return D;

106     }

107 }

108 

109 int main()

110 {

111     init();

112     int T, cas = 0;

113     scanf( "%d", &T );

114     while ( T-- )

115     {

116         int n;

117         scanf( "%d", &n );

118         build( 1, n, 1 );

119 

120         int Q;

121         scanf( "%d", &Q );

122         printf( "Case #%d:\n", ++cas );

123         while ( Q-- )

124         {

125             int a, b;

126             scanf( "%d%d", &a, &b );

127             node ans = Query( a, b, 1, n, 1 );

128 

129             int cnt = 0;

130             bool first = false;

131             for ( int i = 9; i >= 0 && cnt < 5; --i )

132                 if ( ans.root & ( 1 << i ) )

133                 {

134                     if ( first ) putchar(' ');

135                     printf( "%d", i );

136                     first = true;

137                     ++cnt;

138                 }

139             for ( ; cnt < 5; ++cnt ) printf(" -1");

140             puts("");

141         }

142 

143         if ( T ) puts("");

144     }

145     return 0;

146 }