FZU2105-Digits Count(线段树区间）

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

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.

                                                                                   

四种操作：且，或，异或，求和

线段树区间操作。

各种坑点，WA了n多遍~

1.mat 数组没有清零 

2.且 和 或其实是同一种操作，set 1,和 set 0，异或相当于加的操作。所以且 和 或的操作 优先于异或。

3.在AND 处 v=0也是有效操作。

CODE：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <string>
#include <cstring>
#include <queue>
#include <stack>
#include <vector>
#include <set>
#include <map>
const int inf=0xfffffff;
typedef long long ll;
using namespace std;

const int Max=1000006;
int mat[4][Max];

void change(int n,int ii)
{
int k=0;
while(n>0){
mat[k++][ii]=n%2;
n/=2;
}
}
struct segmentree
{
int sum[Max<<2];
int seto[Max<<2];
int setx[Max<<2];
void pushup(int pos){
sum[pos]=sum[pos*2]+sum[pos*2+1];
}
void clean(int pos,int l,int r,int ii){
seto[pos]=-1;
setx[pos]=0;
if(l==r) sum[pos]=0;
else{
int mid=(l+r)/2;
clean(pos*2,l,mid,ii);
clean(pos*2+1,mid+1,r,ii);
pushup(pos);
}
}
void build(int pos,int l,int r,int ii){
if(l==r) sum[pos]=mat[ii][l];
else{
int mid=(l+r)/2;
build(pos*2,l,mid,ii);
build(pos*2+1,mid+1,r,ii);
pushup(pos);
}
}
void pushdown_xor(int pos,int l,int r){
if(setx[pos]!=0){
setx[pos*2+1]^=1;
setx[pos*2]^=1;
int mid=(l+r)/2;
sum[pos*2]=(mid-l+1)-sum[pos*2];
sum[pos*2+1]=(r-mid)-sum[pos*2+1];
setx[pos]=0;
//seto[pos*2]=seto[pos*2+1]=-1;
}
}
void pushdown_or_and(int pos,int l,int r){
if(seto[pos]!=-1){
seto[pos*2]=seto[pos*2+1]=seto[pos];
int mid=(l+r)/2;
sum[pos*2]=(mid-l+1)*seto[pos*2];
sum[pos*2+1]=(r-mid)*seto[pos*2+1];
seto[pos]=-1;
setx[pos*2]=setx[pos*2+1]=0;
}
}
void update_and(int pos,int l,int r,int x,int y){
if(x<=l && y>=r){
seto[pos]=0;
sum[pos]=0;
setx[pos]=0;
}
else{
pushdown_or_and(pos,l,r);
pushdown_xor(pos,l,r);
int mid=(l+r)/2;
if(x<=mid) update_and(pos*2,l,mid,x,y);
if(y>mid) update_and(pos*2+1,mid+1,r,x,y);
pushup(pos);
}
}
void update_or(int pos,int l,int r,int x,int y){
if(x<=l && y>=r){
seto[pos]=1;
sum[pos]=(r-l+1)*seto[pos];
setx[pos]=0;
}
else{
pushdown_or_and(pos,l,r);
pushdown_xor(pos,l,r);
int mid=(l+r)/2;
if(x<=mid) update_or(pos*2,l,mid,x,y);
if(y>mid) update_or(pos*2+1,mid+1,r,x,y);
pushup(pos);
}
}
void update_xor(int pos,int l,int r,int x,int y){
if(x<=l && y>=r){
setx[pos]^=1;
sum[pos]=(r-l+1)-sum[pos];
//seto[pos]=-1;
}
else{
pushdown_or_and(pos,l,r);
pushdown_xor(pos,l,r);
int mid=(l+r)/2;
if(x<=mid) update_xor(pos*2,l,mid,x,y);
if(y>mid) update_xor(pos*2+1,mid+1,r,x,y);
pushup(pos);
}
}
int query(int pos,int l,int r,int x,int y,int ii){
if(x<=l && y>=r) return sum[pos]*(1<<ii);
else{
int ans=0;
pushdown_or_and(pos,l,r);
pushdown_xor(pos,l,r);
int mid=(l+r)/2;
if(x<=mid) ans+=query(pos*2,l,mid,x,y,ii);
if(y>mid) ans+=query(pos*2+1,mid+1,r,x,y,ii);
return ans;
}
}
}tree[4];;

int main()
{
//freopen("in.in","r",stdin);
int T;
scanf("%d",&T);
while(T--){
memset(mat,0,sizeof(mat));
int N,M;
scanf("%d %d",&N,&M);

for(int i=0;i<4;i++){
tree[i].clean(1,1,N,i);
}

for(int i=1;i<=N;i++){
int n;
scanf("%d",&n);
change(n,i);
}
for(int i=0;i<4;i++){
tree[i].build(1,1,N,i);
}
char s[10];

while(M--){
getchar();
int v,xx,yy;
scanf("%s",s);
if(s[0]=='S'){
scanf("%d%d",&xx,&yy);
xx++;yy++;
int ans=0;
for(int i=0;i<4;i++)
ans+=tree[i].query(1,1,N,xx,yy,i);
printf("%d\n",ans);
}
else{
scanf("%d%d%d",&v,&xx,&yy);
xx++;yy++;
if(s[0]=='A'){
int k=0;
while(k<4){
if(v%2==0) tree[k].update_and(1,1,N,xx,yy);
v/=2;
k++;
}
}
if(s[0]=='X'){
int k=0;
while(k<4){
if(v%2) tree[k].update_xor(1,1,N,xx,yy);
v/=2;
k++;
}
}
if(s[0]=='O'){
int k=0;
while(k<4){
if(v%2) tree[k].update_or(1,1,N,xx,yy);
v/=2;
k++;
}
}
}
}
}
return 0;
}
感谢BMan 帮我debug>.<