FZU 2105 Digits Count
Time Limit:10000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u
Practice
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
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.
ac:
#include"algorithm"
#include"iostream"
#include"cstring"
#include"cstdlib"
#include"cstdio"
#include"string"
#include"vector"
#include"stack"
#include"queue"
#include"cmath"
#include"map"
using namespace std;
typedef long long LL ;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define FK(x) cout<<"["<>1);
sum[rt<<1|1][i]=m>>1;
}
if(lazy[rt][i]==2) { //如果进行了XOR操作
if(lazy[rt<<1][i]==INF) { //如果没有进行过任何操作，标记为XOR操作
lazy[rt<<1][i]=2;
sum[rt<<1][i]=m-(m>>1)-sum[rt<<1][i];
} else if(lazy[rt<<1][i]==2) { //如果进行过XOR操作，a^b^b==a 恢复操作内容。
lazy[rt<<1][i]=INF;
sum[rt<<1][i]=m-(m>>1)-sum[rt<<1][i];
} else { //如果进行了操作并且不是XOR操作 将该操作再取XOR操作
lazy[rt<<1][i]^=1;
if(lazy[rt<<1][i]==0) sum[rt<<1][i]=0;
else sum[rt<<1][i]=m-(m>>1);
}
// 另一棵子树用同样的方法处理
if(lazy[rt<<1|1][i]==INF) {
lazy[rt<<1|1][i]=2;
sum[rt<<1|1][i]=(m>>1)-sum[rt<<1|1][i];
} else if(lazy[rt<<1|1][i]==2) {
lazy[rt<<1|1][i]=INF;
sum[rt<<1|1][i]=(m>>1)-sum[rt<<1|1][i];
} else {
lazy[rt<<1|1][i]^=1;
if(lazy[rt<<1|1][i]==0) sum[rt<<1|1][i]=0;
else sum[rt<<1|1][i]=(m>>1);
}
}
lazy[rt][i]=INF; //标记lazy为空
}
void Build(int l,int r,int rt) {
for(int i=0; i<4; i++) lazy[rt][i]=INF; //清空懒惰标记
if(r==l) {
int temp;
scanf("%d",&temp);
// FK("temp=="<>1;
Build(lson);
Build(rson);
for(int i=0; i<4; i++) PushUp(rt,i);
}
void UpData(int L,int R,int v,int i,int l,int r,int rt) {
if(r<=R&&L<=l) {
switch(v) {
case 0:
sum[rt][i]=0,lazy[rt][i]=v;
//如果是进行AND操作，并且是0，清空和。
break;
case 1:
sum[rt][i]=r-l+1,lazy[rt][i]=v;
//如果是进行OR 操作，并且是1，填满和。
break;
case 2:
sum[rt][i]=r-l+1-sum[rt][i];
if(lazy[rt][i]==2) lazy[rt][i]=INF;
else if(lazy[rt][i]==INF) lazy[rt][i]=2;
else lazy[rt][i]^=1;
break;
default:
break;
}
return ;
}
PushDown(rt,r-l+1,i);
int m=(r+l)>>1;
if(L<=m)UpData(L,R,v,i,lson);
if(R>m) UpData(L,R,v,i,rson);
PushUp(rt,i);
}
int Query(int L,int R,int i,int l,int r,int rt) {
if(L<=l&&r<=R) {
return sum[rt][i];
// 返回这个数该位的和。
}
int m=(r+l)>>1;
int sum=0;
PushDown(rt,r-l+1,i);
if(L<=m)sum+=Query(L,R,i,lson);
if(R>m) sum+=Query(L,R,i,rson);
return sum;
}
int main() {
int T;
scanf("%d",&T);
bigfor(T) {
int n,m;
scanf("%d%d",&n,&m);
char op[5];
Build(0,n-1,1);
// FK("Build Success!");
for(int i=0; i1,1->0【区间更新】
for(int j=0; j<4; j++) {
int x=opn&(1<
