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<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

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<A[i]<16)的序列,一共有4种操作:①AND opn L R,将区间[L,R]内的数与opn进行与运算;②OR opn L R,将区间[L,R]内的数与opn进行或运算
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<cstdio>#include<algorithm>#include<string.h>#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1using 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<<i)) sum[i][rt]=1;//将序列中每个数的每一位存入线段树        return;    }    int m=(l+r)>>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<<i)){//当第i位是1的时候进行或/异或运算                int len=r-l+1;                if(op=='O'){                    sum[i][rt]=len;                    cover[i][rt]=1;                    X[i][rt]=0;                }                else if(op=='X'){                    sum[i][rt]=len-sum[i][rt];                    FXOR(i,rt);                }            }            else if(op=='A'){//当第i位是0的时候进行与运算                sum[i][rt]=0;                cover[i][rt]=0;                X[i][rt]=0;            }        }        return;    }    PushDown(rt,r-l+1);    int m=(l+r)>>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<<i);//总和=第i位1的个数*2^i        return ans;    }    PushDown(rt,r-l+1);    int m=(l+r)>>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;}