FZU-2105 Digits Count(线段树)

Problem 2105 Digits Count

Accept: 441    Submit: 2070
Time Limit: 10000 mSec    Memory Limit : 262144 KB

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

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

14 41 2 4 7SUM 0 2XOR 5 0 0OR 6 0 3SUM 0 2

Sample Output

718

Hint

A = [1 2 4 7]

SUM 0 2, result=1+2+4=7;

XOR 5 0 0, A=[4 2 4 7];

OR 6 0 3, A=[6 6 6 7];

SUM 0 2, result=6+6+6=18.

Source

“高教社杯”第三届福建省大学生程序设计竞赛 

题意:给出有n个数(0 ③XOR opn L R,将区间[L,R]内的数与opn进行异或运算;④SUM L R,输出区间[L,R]内的数之和.(0<=opn<=15)

题解:位运算的线段树,我们发现A[i]和opn都是小于16的,就说明它们的二进制位数不超过4,那么我们可以将这些数的每一位用线段树保存,然后在每一位上进行位运算,
这样就可以起到线段树的维护效果.下面对于每种运算进行分析:①与运算:1&1=1,1&0=0,0&0=0,从这3个式子可以看出,进行与运算的时候,如果opn的某一位是0,那么A[i]
相应的那一位就一定要变成0,然而opn的某一位是1,进行与运算之后A[i]的相应为还是不会改变的;②或运算:1|1=1,1|0=1,0|0=0,其实或运算和与运算就是完全相反的,
如果opn的某一位是1,那么A[i]相应的哪一位就一定要变成1,然而opn的某一位是0,进行与运算之后A[i]的相应为还是不会改变的;③异或运算:1^1=0,1^0=1,0^0=0,其实这
道题最难的就是异或运算,前两种运算都是直接覆盖这段区间就可以了,然而异或运算时反转,这样的话就需要再加一个懒惰标记X,跟或运算相同的是,只有当某一位是1的时候
才要进行异或运算,如果这一位被标记了覆盖的懒惰标记(cover[rt]!=-1),那么就将他反转(oover[rt]^=1),因为如果这一位是要覆盖为1的,反转之后自然就是覆盖为0,如果
没有标记覆盖的懒惰标记,则将反转的懒惰标记进行反转(X[rt]^=1);

#include
#include
#include
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
using namespace std;
typedef long long LL;
const int maxn = 1e6 + 5;
int sum[4][maxn<<2];
int cover[4][maxn<<2],X[4][maxn<<2];
void PushUP(int rt){
    for(int i=0;i<4;i++) sum[i][rt]=sum[i][rt<<1]+sum[i][rt<<1|1];
}
void build(int l,int r,int rt){
    for(int i=0;i<4;i++){
        cover[i][rt]=-1;//覆盖标记初始化
        X[i][rt]=0;//反转标记初始化
        sum[i][rt]=0;//总和初始化
    }
    if(l==r){
        int x;
        scanf("%d",&x);
        for(int i=0;i<4;i++) if(x&(1<>1;
    build(lson);
    build(rson);
    PushUP(rt);
}
void FXOR(int i,int rt){//进行反转操作
    if(cover[i][rt]!=-1) cover[i][rt]^=1;
    else  X[i][rt]^=1;
}
void PushDown(int rt,int m){//将懒惰标记下移
    for(int i=0;i<4;i++) {
        if(cover[i][rt]!=-1){//如果有覆盖标记就把覆盖标记下移
            sum[i][rt<<1]=(m-(m>>1))*cover[i][rt];
            sum[i][rt<<1|1]=(m>>1)*cover[i][rt];
            cover[i][rt<<1]=cover[i][rt<<1|1]=cover[i][rt];
            X[i][rt<<1]=X[i][rt<<1|1]=0;
            cover[i][rt]=-1;
        }
        if(X[i][rt]){//如果有反转标记就把反转标记下移
            sum[i][rt<<1]=m-(m>>1)-sum[i][rt<<1];
            sum[i][rt<<1|1]=(m>>1)-sum[i][rt<<1|1];
            FXOR(i,rt<<1);
            FXOR(i,rt<<1|1);
            X[i][rt]=0;
        }
    }
}
void update(char op,int L,int R,int c,int l,int r,int rt){
    if(L<=l&&R>=r){
        for(int i=0;i<4;i++){
            if(c&(1<>1;
    if(L<=m) update(op,L,R,c,lson);
    if(R>m) update(op,L,R,c,rson);
    PushUP(rt);
}
int query(int L,int R,int l,int r,int rt){
    if(L<=l&&R>=r){
        int ans=0;
        for(int i=0;i<4;i++) ans+=sum[i][rt]*(1<>1;
    int ret=0;
    if(L<=m) ret+=query(L,R,lson);
    if(R>m) ret+=query(L,R,rson);
    PushUP(rt);
    return ret;
}
int main(){
    int T;
   // freopen("input.txt","r",stdin);
    scanf("%d",&T);
    while(T--){
        int n,m;
        scanf("%d%d",&n,&m);
        build(0,n-1,1);
        int a,b,c;
        char op[15];
        while(m--){
            scanf("%s",op);
            if(op[0]=='S'){
                scanf("%d%d",&a,&b);
                printf("%d\n",query(a,b,0,n-1,1));
            }
            else {
                scanf("%d%d%d",&c,&a,&b);
                update(op[0],a,b,c,0,n-1,1);
            }
        }
    }
    return 0;
}