FZU 2105 (线段树)

 Problem 2105 Digits Count

 Problem Description

Given N integers A={A[0],A[1],...,A[N-1]}. Here we have some operations:

Operation 1: AND opn L R

Here opn, L and R are integers.

For L≤i≤R, we do A[i]=A[i] AND opn (here "AND" is bitwise operation).

Operation 2: OR opn L R

Here opn, L and R are integers.

For L≤i≤R, we do A[i]=A[i] OR opn (here "OR" is bitwise operation).

Operation 3: XOR opn L R

Here opn, L and R are integers.

For L≤i≤R, we do A[i]=A[i] XOR opn (here "XOR" is bitwise operation).

Operation 4: SUM L R

We want to know the result of A[L]+A[L+1]+...+A[R].

Now can you solve this easy problem?

 Input

The first line of the input contains an integer T, indicating the number of test cases. (T≤100)

Then T cases, for any case, the first line has two integers n and m (1≤n≤1,000,000, 1≤m≤100,000), indicating the number of elements in A and the number of operations.

Then one line follows n integers A[0], A[1], ..., A[n-1] (0≤A[i]<16,0≤i<n).

Then m lines, each line must be one of the 4 operations above. (0≤opn≤15)

 Output

For each test case and for each "SUM" operation, please output the result with a single line.

 Sample Input

1 4 4 1 2 4 7 SUM 0 2 XOR 5 0 0 OR 6 0 3 SUM 0 2

 Sample Output

7 18

 题意不多说了。 观察下a的值表示成二进制不会超过4位内存刚刚够。对每一位维护一下线段树就好了。

具体维护方法如下：

由于  :   1 & 0 = 0 

     0 & 0 =0    所以&0会改变区间值。

1& 1 =1

  0&1=0  所以&1 区间值不变，可以忽略。

同理可以分析其他的操作。然后线段树lazy维护一下1的个数就好了。  注意xor 和& or是互斥的，也就是说当标记了&或or时应把标记 xor清空。

  1 // by cao ni ma
  2 // hehe
  3 #include <cstdio>
  4 #include <cstring>
  5 #include <algorithm>
  6 #include <vector>
  7 #include <queue>
  8 using namespace std;
  9 const int MAX = 1000000+5;
 10 typedef long long ll;
 11 int sum[4][MAX<<2];
 12 int col_or[4][MAX<<2],col_xor[4][MAX<<2];
 13 int A[MAX];
 14 void pushup(int o,int cur){
 15     sum[cur][o]=sum[cur][o<<1]+sum[cur][o<<1|1];
 16 }
 17 
 18 void pushdown(int o,int cur,int m){
 19     if(col_or[cur][o]!=-1){
 20         col_xor[cur][o<<1]=col_xor[cur][o<<1|1]=0;
 21         col_or[cur][o<<1]=col_or[cur][o<<1|1]=col_or[cur][o];
 22         sum[cur][o<<1]=(m-(m>>1))*col_or[cur][o<<1];
 23         sum[cur][o<<1|1]=(m>>1)*col_or[cur][o<<1|1];
 24         col_or[cur][o]=-1;
 25     }
 26     if(col_xor[cur][o]){
 27         col_xor[cur][o<<1]^=1,col_xor[cur][o<<1|1]^=1;
 28         sum[cur][o<<1]=((m-(m>>1))-sum[cur][o<<1]);
 29         sum[cur][o<<1|1]=((m>>1)-sum[cur][o<<1|1]);
 30         col_xor[cur][o]=0;
 31     }
 32 }
 33 
 34 void build(int L,int R,int o,int cur){
 35     col_or[cur][o]=-1;
 36     col_xor[cur][o]=0;
 37     if(L==R){
 38         sum[cur][o]=((A[L]&(1<<cur))?1:0);
 39     }
 40     else{
 41         int mid=(L+R)>>1;
 42         build(L,mid,o<<1,cur);
 43         build(mid+1,R,o<<1|1,cur);
 44         pushup(o,cur);
 45     }
 46 }
 47 
 48 void modify2(int L,int R,int o,int ls,int rs,int v,int cur){
 49     if(ls<=L && rs>=R){
 50         col_xor[cur][o]=0;
 51         col_or[cur][o]=v;
 52         sum[cur][o]=v*(R-L+1);
 53         return ;
 54     }
 55     pushdown(o,cur,R-L+1);
 56     int mid=(R+L)>>1;
 57     if(ls<=mid) modify2(L,mid,o<<1,ls,rs,v,cur);
 58     if(rs>mid) modify2(mid+1,R,o<<1|1,ls,rs,v,cur);
 59     pushup(o,cur);
 60 
 61 }
 62 
 63 void modify1(int L,int R,int o,int ls,int rs,int v,int cur){
 64     if(ls<=L && rs>=R){
 65         if(col_or[cur][o]!=-1){
 66             col_or[cur][o]^=1;
 67             sum[cur][o]=(R-L+1)-sum[cur][o];
 68             return ;
 69         }
 70         else{
 71             col_xor[cur][o]^=1;
 72             sum[cur][o]=(R-L+1)-sum[cur][o];
 73             return ;
 74         }
 75     }
 76     pushdown(o,cur,R-L+1);
 77     int mid=(R+L)>>1;
 78     if(ls<=mid) modify1(L,mid,o<<1,ls,rs,v,cur);
 79     if(rs>mid) modify1(mid+1,R,o<<1|1,ls,rs,v,cur);
 80     pushup(o,cur);
 81 }
 82 
 83 int Query(int L,int R,int o,int ls,int rs,int cur) {
 84     if(ls<=L && rs>=R) return sum[cur][o];
 85     pushdown(o,cur,R-L+1);
 86     int mid=(R+L)>>1; int ans=0;
 87     if(ls<=mid) ans+=Query(L,mid,o<<1,ls,rs,cur);
 88     if(rs>mid) ans+=Query(mid+1,R,o<<1|1,ls,rs,cur);
 89     return ans;
 90 }
 91 
 92 int main(){
 93     int n,m,cas,ls,rs,val;
 94     char op[6];
 95     scanf("%d",&cas);
 96     while(cas--){
 97         scanf("%d %d",&n,&m);
 98         for(int i=1;i<=n;i++) {
 99             scanf("%d",&A[i]);
100         }
101         for(int i=0;i<4;i++) {
102             build(1,n,1,i);
103         }
104         for(int i=0;i<m;i++) {
105             scanf("%s",op);
106             if(op[0]=='S'){
107                 scanf("%d %d",&ls,&rs);
108                 ls++,rs++;
109                 int ans=0;
110                 for(int i=0;i<4;i++) {
111                     ans+=Query(1,n,1,ls,rs,i)*(1<<i);
112                 }
113                 printf("%d
",ans);
114             }
115             else if(op[0]=='O'){
116                 scanf("%d %d %d",&val,&ls,&rs);
117                 rs++,ls++;
118                 for(int i=0;i<4;i++) {
119                     if(val&(1<<i)){
120                         modify2(1,n,1,ls,rs,1,i);
121                     }
122                 }
123             }
124             else if(op[0]=='A'){
125                 scanf("%d %d %d",&val,&ls,&rs);
126                 rs++,ls++;
127                 for(int i=0;i<4;i++) {
128                     if(!(val&(1<<i))){
129                        modify2(1,n,1,ls,rs,0,i);
130                     }
131                 }
132             }
133             else{
134                 scanf("%d %d %d",&val,&ls,&rs);
135                 rs++,ls++;
136                 for(int i=0;i<4;i++) {
137                     if((val&(1<<i))){
138                          modify1(1,n,1,ls,rs,1,i);
139                     }
140                 }
141             }
142         }
143     }
144     return 0;

145 }