FZU 2105 Digits Count(线段树)

Problem 2105 Digits Count

Accept: 302 Submit: 1477

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≤in).

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.

由于数据特别多,但是数据的值不大(<16),所以必有大量重复的元素就可以进行区间合并

#include <set>
#include <map>
#include <list>
#include <stack>
#include <cmath>
#include <vector>
#include <queue>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#define PI cos(-1.0)
#define RR freopen("input.txt","r",stdin)
using namespace std;
typedef long long LL;
const int MAX = 1000010;
int Seg[MAX*6];
int Arr[MAX];
int n,m;
int Oper(int num,int PN,int OP)//进行操作
{
switch(OP)
{
case 1:
return num&PN;
case 2:
return num|PN;
case 3:
return num^PN;
}
return 0;
}

void Build(int L,int R,int site)//建立线段树
{
if(L==R)
{
Seg[site]=Arr[L];
return ;
}
int mid=(L+R)>>1;
Build(L,mid,site<<1);
Build(mid+1,R,site<<1|1);
if(Seg[site<<1]==Seg[site<<1|1]&&Seg[site<<1]!=1)//将值相同的区间进行合并,如果不相同则为-1.
{
Seg[site]=Seg[site<<1];
}
else
{
Seg[site]=-1;
}
}
void Update(int L,int R,int l,int r,int site,int PN,int OP)//更新操作
{
if(L==l&&R==r&&Seg[site]!=-1)//三种操作
{
Seg[site]=Oper(Seg[site],PN,OP);
return ;
}
int mid = (L+R)>>1;
if(Seg[site]!=-1)//向下更新,如过对已经合并的区间里面进行操作,则需要将区间先拆分,更行完以后再判断是否可以合并
{
Seg[site<<1]=Seg[site<<1|1]=Seg[site];
Seg[site]=-1;
}
if(r<=mid)
{
Update(L,mid,l,r,site<<1,PN,OP);
}
else if(l>mid)
{
Update(mid+1,R,l,r,site<<1|1,PN,OP);
}
else
{
Update(L,mid,l,mid,site<<1,PN,OP);
Update(mid+1,R,mid+1,r,site<<1|1,PN,OP);
}
if(Seg[site<<1]==Seg[site<<1|1]&&Seg[site<<1]!=1)//区间合并
{
Seg[site]=Seg[site<<1];
}
}
int Query(int L,int R,int l,int r,int site)//查询
{
if(L==l&&R==r&&Seg[site]!=-1)
{
return (R-L+1)*Seg[site];
}
if(Seg[site]!=-1)//查询的时候,如果要查询一个区间内的区间,则需要先将区间向下更新(想想为什么?);
{
Seg[site<<1]=Seg[site<<1|1]=Seg[site];
Seg[site]=-1;
}
int mid=(L+R)>>1;
if(r<=mid)
{
return Query(L,mid,l,r,site<<1);
}
else if(l>mid)
{
return Query(mid+1,R,l,r,site<<1|1);
}
else
{
return Query(L,mid,l,mid,site<<1)+Query(mid+1,R,mid+1,r,site<<1|1);
}
}
int main()
{
int T;
char str[15];
int l,r,PN;
scanf("%d",&T);
while(T--)
{
scanf("%d %d",&n,&m);
for(int i=0;i<n;i++)
{
scanf("%d",&Arr[i]);
}
Build(0,n-1,1);
for(int i=1;i<=m;i++)
{
scanf("%s",str);
if(strcmp(str,"SUM")==0)
{
scanf("%d %d",&l,&r);
printf("%d\n",Query(0,n-1,l,r,1));
}
else
{
scanf("%d %d %d",&PN,&l,&r);
if(strcmp(str,"AND")==0)
{
Update(0,n-1,l,r,1,PN,1);
}
else if(strcmp(str,"OR")==0)
{
Update(0,n-1,l,r,1,PN,2);
}
else if(strcmp(str,"XOR")==0)
{
Update(0,n-1,l,r,1,PN,3);
}
}
}
}
return 0;
}