qwq ， ylx 问我要一份线段树的版 ， 可我线段树一直是10分钟 ，从不写版 ，qwq ，还是放一份版在这 。
题目见 ： http://acm.hdu.edu.cn/showproblem.php?pid=1166
只是单点查询+区间修改（zkw版本） ，唯一注意的就是数组不能开小了（我先就是数组开小了 ，一直错）
1 #include <iostream> 2 #include <cstdlib> 3 #include <cstring> 4 #include <cstdio> 5 const int inf = 1<<30 , maxn = 1 << 16 ; 6 int t , n , M , tree[maxn<<1] ; 7 char ch[15] ; 8 using namespace std ; 9 10 void build( int num ) 11 { 12 for( M = 1 ; M <= num + 1 ; M <<= 1 ) ; 13 for( int i = M + 1 ; i <= M + num ; ++i ) 14 scanf( "%d" , tree+i ) ; 15 for( int i = M - 1 ; i > 0 ; --i ) 16 tree[i] = tree[i<<1] + tree[i<<1|1] ; 17 } 18 19 20 21 22 void update( int num , int add ) 23 { 24 for( tree[num+=M] += add , num>>=1 ; num; num>>=1 ) 25 tree[num] = tree[num<<1] + tree[num<<1|1] ; 26 } 27 28 29 int quer( int l , int r ) 30 { 31 int ans = 0 ; 32 for( l = l + M - 1 , r = r + M + 1 ; l^r^1 ; l >>= 1 , r >>= 1 ) 33 { 34 if( ~l&1 ) ans += tree[l^1] ; 35 if( r & 1 ) ans += tree[r^1] ; 36 } 37 return ans ; 38 } 39 40 int main( ) 41 { 42 scanf( "%d" , &t ) ; 43 int a , b , i = 0 ; 44 while( t-- ) 45 { 46 scanf( "%d" , &n ) ; 47 memset( tree, 0, sizeof(tree) ); 48 build( n ) ; 49 printf("Case %d: " , ++i ); 50 while( 1 ) 51 { 52 53 scanf( "%s" , ch ) ; 54 if( ch[0] == 'E' ) break ; 55 scanf( "%d%d" , &a , &b ) ; 56 if( ch[0] == 'Q' ) 57 { 58 printf( "%d " , quer( a , b ) ) ; 59 } 60 else 61 { 62 if( ch[0] == 'S' ) b = -b ; 63 update( a , b ) ; 64 } 65 } 66 } 67 return 0 ; 68 }
题目见：http://poj.org/problem?id=3468
这是区间修改+区间查询普通版
1 #include <iostream> 2 #include <cstring> 3 #include <cstdlib> 4 #include <cstdio> 5 const int inf = 1<<30 , maxn = 100000 + 11 ; 6 using namespace std; 7 int n , q , cnt ; 8 struct id 9 { 10 long long sum , lazy ; 11 }tree[maxn<<2] ; 12 13 long long int read( ) { 14 char ch = getchar( ) ; long long int ret = 0 , k = 1 ; 15 while( ch < '0' || ch > '9' ) { if( ch == '-' ) k = -1 ; ch = getchar( ) ; } 16 while( ch >= '0' && ch <= '9' ) ret = ret * 10 + ch - '0' , ch = getchar( ) ; 17 return ret * k ; 18 } 19 20 21 22 void set_lazy( int num , int l , int r ) 23 { 24 int m = r - l + 1 ; 25 if( tree[num].lazy ) 26 { 27 tree[num<<1].lazy += tree[num].lazy ; 28 tree[(num<<1)+1].lazy += tree[num].lazy ; 29 tree[num<<1].sum += tree[num].lazy * ( m - (m>>1) ) ; 30 tree[(num<<1)+1].sum += tree[num].lazy * ( m>>1 ) ; 31 tree[num].lazy = 0ll ; 32 } 33 34 } 35 36 void update( int l , int r , int num , int L , int R , int add ) 37 { 38 if( l == L && r == R ) 39 { 40 tree[num].sum += 1ll*add * ( R - L + 1 ) ; 41 tree[num].lazy += add ; 42 return ; 43 } 44 if( l == r ) return ; 45 set_lazy( num , l , r ) ; 46 int mid = l + ( ( r - l ) >> 1 ); 47 if( R <= mid ) update( l , mid , num<<1 , L , R , add ) ; 48 else if( L > mid ) update( mid+1 , r , (num<<1)+1 , L , R , add ) ; 49 else 50 { 51 update( l , mid , num<<1 , L , mid , add ) ; 52 update( mid+1 , r , (num<<1)+1 , mid+1 , R , add ) ; 53 } 54 tree[num].sum = tree[num<<1].sum + tree[(num<<1)+1].sum ; 55 } 56 57 58 long long int quer( int l , int r , int num , int L , int R ) 59 { 60 if( l == L && R == r ) return tree[num].sum ; 61 set_lazy( num , l , r ) ; 62 int mid = l + ( ( r - l ) >> 1 ) ; 63 if( R <= mid ) return quer( l , mid , num<<1 , L , R ) ; 64 else if( L > mid ) return quer( mid + 1 , r , (num<<1)+1 , L , R ) ; 65 else return quer( l , mid , num<<1 , L , mid ) + quer( mid+1 , r , (num<<1)+1 , mid+1 , R ) ; 66 } 67 68 69 void build( int num , int l , int r ) 70 { 71 tree[num].lazy = 0ll ; 72 if( l == r ) 73 { 74 tree[num].sum = read( ) ; 75 return ; 76 } 77 int mid = l + ( ( r - l ) >> 1 ) ; 78 79 build( num<<1 , l , mid ) ; 80 build( (num<<1)+1 , mid+1 , r ) ; 81 tree[num].sum = tree[num<<1].sum + tree[(num<<1)+1].sum ; 82 } 83 84 int main( ) 85 { 86 n = read( ) , q = read( ) ; 87 build( 1 , 1 , n ) ; 88 for( int x = 1 ; x <= q ; ++x ) 89 { 90 char a ; int b , c , d ; 91 scanf( "%s" , &a ) ; 92 b = read( ) , c = read( ) ; 93 if( a == 'C' ) 94 { 95 d = read( ) ; 96 update( 1 , n , 1 , b , c , d ) ; 97 } 98 if( a == 'Q' ) 99 { 100 printf( "%lld " , quer( 1 , n , 1 , b , c ) ) ; 101 } 102 } 103 }