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 = +;
 10 typedef long long ll;
 11 int sum[][MAX<<];
 12 int col_or[][MAX<<],col_xor[][MAX<<];
 13 int A[MAX];
 14 void pushup(int o,int cur){
 15     sum[cur][o]=sum[cur][o<<]+sum[cur][o<<|];
 16 }
 17 
 18 void pushdown(int o,int cur,int m){
 19     if(col_or[cur][o]!=-){
 20         col_xor[cur][o<<]=col_xor[cur][o<<|]=;
 21         col_or[cur][o<<]=col_or[cur][o<<|]=col_or[cur][o];
 22         sum[cur][o<<]=(m-(m>>))*col_or[cur][o<<];
 23         sum[cur][o<<|]=(m>>)*col_or[cur][o<<|];
 24         col_or[cur][o]=-;
 25     }
 26     if(col_xor[cur][o]){
 27         col_xor[cur][o<<]^=,col_xor[cur][o<<|]^=;
 28         sum[cur][o<<]=((m-(m>>))-sum[cur][o<<]);
 29         sum[cur][o<<|]=((m>>)-sum[cur][o<<|]);
 30         col_xor[cur][o]=;
 31     }
 32 }
 33 
 34 void build(int L,int R,int o,int cur){
 35     col_or[cur][o]=-;
 36     col_xor[cur][o]=;
 37     if(L==R){
 38         sum[cur][o]=((A[L]&(<<cur))?:);
 39     }
 40     else{
 41         int mid=(L+R)>>;
 42         build(L,mid,o<<,cur);
 43         build(mid+,R,o<<|,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]=;
 51         col_or[cur][o]=v;
 52         sum[cur][o]=v*(R-L+);
 53         return ;
 54     }
 55     pushdown(o,cur,R-L+);
 56     int mid=(R+L)>>;
 57     if(ls<=mid) modify2(L,mid,o<<,ls,rs,v,cur);
 58     if(rs>mid) modify2(mid+,R,o<<|,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]!=-){
 66             col_or[cur][o]^=;
 67             sum[cur][o]=(R-L+)-sum[cur][o];
 68             return ;
 69         }
 70         else{
 71             col_xor[cur][o]^=;
 72             sum[cur][o]=(R-L+)-sum[cur][o];
 73             return ;
 74         }
 75     }
 76     pushdown(o,cur,R-L+);
 77     int mid=(R+L)>>;
 78     if(ls<=mid) modify1(L,mid,o<<,ls,rs,v,cur);
 79     if(rs>mid) modify1(mid+,R,o<<|,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+);
 86     int mid=(R+L)>>; int ans=;
 87     if(ls<=mid) ans+=Query(L,mid,o<<,ls,rs,cur);
 88     if(rs>mid) ans+=Query(mid+,R,o<<|,ls,rs,cur);
 89     return ans;
 90 }
 91 
 92 int main(){
 93     int n,m,cas,ls,rs,val;
 94     char op[];
 95     scanf("%d",&cas);
 96     while(cas--){
 97         scanf("%d %d",&n,&m);
 98         for(int i=;i<=n;i++) {
 99             scanf("%d",&A[i]);
         }
         for(int i=;i<;i++) {
             build(,n,,i);
         }
         for(int i=;i<m;i++) {
             scanf("%s",op);
             if(op[]=='S'){
                 scanf("%d %d",&ls,&rs);
                 ls++,rs++;
                 int ans=;
                 for(int i=;i<;i++) {
                     ans+=Query(,n,,ls,rs,i)*(<<i);
                 }
                 printf("%d\n",ans);
             }
             else if(op[]=='O'){
                 scanf("%d %d %d",&val,&ls,&rs);
                 rs++,ls++;
                 for(int i=;i<;i++) {
                     if(val&(<<i)){
                         modify2(,n,,ls,rs,,i);
                     }
                 }
             }
             else if(op[]=='A'){
                 scanf("%d %d %d",&val,&ls,&rs);
                 rs++,ls++;
                 for(int i=;i<;i++) {
                     if(!(val&(<<i))){
                        modify2(,n,,ls,rs,,i);
                     }
                 }
             }
             else{
                 scanf("%d %d %d",&val,&ls,&rs);
                 rs++,ls++;
                 for(int i=;i<;i++) {
                     if((val&(<<i))){
                          modify1(,n,,ls,rs,,i);
                     }
                 }
             }
         }
     }
     return ;

145 }