FZU 2105 Digits Count
Problem 2105 Digits Count
Accept: 444 Submit: 2139 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.
线段树，由于数组范围只有0到17，所以会有大量重复的块，所以直接线段树暴力来
#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>
using namespace std;
typedef long long int LL;
const int maxn=1e6;
int num[maxn*4+5];
int sum[maxn*4+5];
int n,m;
void pushup(int node)
{
sum[node]=sum[node<<1]+sum[node<<1|1];
if(num[node<<1]==num[node<<1|1]&&num[node<<1]!=-1)
num[node]=num[node<<1];
else
num[node]=-1;
}
void build(int node,int l,int r)
{
if(l==r)
{
scanf("%d",&num[node]);
sum[node]=num[node];
return;
}
int mid=(l+r)>>1;
build(node<<1,l,mid);
build(node<<1|1,mid+1,r);
pushup(node);
}
void update(int node,int l,int r,int L,int R,int tag,int flag)
{
if(L<=l&&r<=R)
{
if(num[node]!=-1)
{
if(tag==1) num[node]&=flag;
else if(tag==2) num[node]|=flag;
else num[node]^=flag;
sum[node]=num[node]*(r-l+1);
return;
}
int mid=(l+r)>>1;
if(L<=mid)
update(node<<1,l,mid,L,R,tag,flag);
if(R>mid)
update(node<<1|1,mid+1,r,L,R,tag,flag);
pushup(node);
return;
}
int mid=(l+r)>>1;
if(num[node]!=-1)
{
sum[node<<1]=num[node]*(mid-l+1);
sum[node<<1|1]=num[node]*(r-mid);
num[node<<1]=num[node<<1|1]=num[node];
}
if(L<=mid) update(node<<1,l,mid,L,R,tag,flag);
if(R>mid) update(node<<1|1,mid+1,r,L,R,tag,flag);
pushup(node);
}
int query(int node,int l,int r,int L,int R)
{
if(L<=l&&r<=R)
{
return sum[node];
}
int mid=(l+r)>>1;
if(num[node]!=-1)
{
sum[node<<1]=num[node]*(mid-l+1);
sum[node<<1|1]=num[node]*(r-mid);
num[node<<1]=num[node<<1|1]=num[node];
}
int ret=0;
if(L<=mid) ret+=query(node<<1,l,mid,L,R);
if(R>mid) ret+=query(node<<1|1,mid+1,r,L,R);
return ret;
}
int main()
{
int t;
scanf("%d",&t);
char a[10];
int x,y,z;
while(t--)
{
scanf("%d%d",&n,&m);
memset(num,-1,sizeof(num));
build(1,1,n);
for(int i=1;i<=m;i++)
{
scanf("%s",a);
if(a[0]=='A')
{
scanf("%d%d%d",&x,&y,&z);
update(1,1,n,y+1,z+1,1,x);
}
else if(a[0]=='O')
{
scanf("%d%d%d",&x,&y,&z);
update(1,1,n,y+1,z+1,2,x);
}
else if(a[0]=='X')
{
scanf("%d%d%d",&x,&y,&z);
update(1,1,n,y+1,z+1,3,x);
}
else
{
scanf("%d%d",&x,&y);
printf("%d\n",query(1,1,n,x+1,y+1));
}
}
}
return 0;
}