FZU-2105 Digits Count (two tagged as segment update)

The main topic: to the number of n 0~15, there are 3 kinds of update operations, 1 kinds of inquiry operation. 3 Kinds of update operations are: 1, let a closed interval of all the numbers and a 0~15 between the number of logic and operation, 2, let a closed interval of all the numbers and a 0~15 between the number of logic or operation; 3, let all the number of a closed interval and a 0~15 between the number of the XOR operation. One type of inquiry is to ask for the sum of all the numbers in a closed interval.

Title Analysis: All the input numbers are between 0~15, each of the binary can establish a line tree, build four trees. There are actually only two of the 3 update operations, namely interval and interval inversion. With two lazy tags, when the interval is updated, to clear the existing reversal markers, so that the mark under the two tags under the sequence of transmission.

The code is as follows:

# include<iostream># include<cstdio># include<cstring># include<algorithm>using namespace std;# define LL long longconst int n=1000000;int n,m;int a[n+5];int tr[4][n*4+5];int lazy[4][n*4+5];int lazy_xor[4][N*4+5 ];char com[5];void pushdown (int id,int rt,int l,int r) {int mid=l+ (r-l)/2;if (lazy[id][rt]!=-1) {lazy_xor[id][rt<< 1]=lazy_xor[id][rt<<1|1]=0;lazy[id][rt<<1]=lazy[id][rt<<1|1]=lazy[id][rt];tr[id][rt<<1 ]=lazy[id][rt]* (mid-l+1); tr[id][rt<<1|1]=lazy[id][rt]* (r-mid); lazy[id][rt]=-1;} int &q=lazy_xor[id][rt];if (q==1) {lazy_xor[id][rt<<1]^=1;lazy_xor[id][rt<<1|1]^=1;tr[id][rt< <1]=mid-l+1-tr[id][rt<<1];tr[id][rt<<1|1]=r-mid-tr[id][rt<<1|1];q=0;}} void pushup (int id,int RT) {tr[id][rt]=tr[id][rt<<1]+tr[id][rt<<1|1];} void build (int id,int rt,int l,int r) {lazy[id][rt]=-1;lazy_xor[id][rt]=0;if (l==r) {if (a[r]& (1<<id)) >0) Tr[id][rt]=1;else tr[id][rt]=0;} Else{int mid=l+ (r-l)/2;Build (Id,rt<<1,l,mid); build (Id,rt<<1|1,mid+1,r);p ushup (ID,RT);}} void update (int id,int rt,int l,int r,int l,int r,int x) {if (l<=l&&r<=r) {if (x==2) {if (lazy[id][rt]!=-1) Lazy[id][rt]^=1;else lazy_xor[id][rt]^=1;tr[id][rt]= (r-l+1)-tr[id][rt];} else{lazy[id][rt]=x;lazy_xor[id][rt]=0;tr[id][rt]=x* (r-l+1);}} Else{pushdown (id,rt,l,r), int mid=l+ (r-l)/2;if (l<=mid) Update (ID,RT&LT;&LT;1,L,MID,L,R,X), if (r>mid) update ( id,rt<<1|1,mid+1,r,l,r,x);p ushup (ID,RT);}} LL query (int id,int rt,int l,int r,int l,int R) {if (l<=l&&r<=r) return TR[ID][RT];p ushdown (id,rt,l,r); int mid=l+ (r-l)/2; LL res=0;if (L<=mid) res+=query (id,rt<<1,l,mid,l,r); if (R>mid) res+=query (id,rt<<1|1,mid+1,r,l,r) ; return res;} int main () {//freopen ("In.txt", "R", stdin), int t;scanf ("%d", &t), while (t--) {scanf ("%d%d", &n,&m); for (int I=0;i<n;++i) scanf ("%d", A+i), for (int i=0;i<4;++i) build (i,1,0,n-1); int O,b,c;while (m--) {scanf ("%s", com); if ( com[0]== ' S ') {scanf ("%d%d", & b,&c); int ans=0;for (int i=0;i<4;++i) {ans+=query (i,1,0,n-1,b,c) * (1<<i);} printf ("%d\n", ans);} ELSE{SCANF ("%d%d%d", &o,&b,&c), if (com[0]== ' X ') {for (int i=0;i<4;++i) if (o& (1<<i)) Update (i , 1,0,n-1,b,c,2);} else if (com[0]== ' O ') {for (int i=0;i<4;++i) if (o& (1<<i)) update (i,1,0,n-1,b,c,1);} else if (com[0]== ' A ') {for (int i=0;i<4;++i) if (! ( o& (1<<i))) update (i,1,0,n-1,b,c,0);}}} return 0;}

　　

FZU-2105 Digits Count (two tagged as segment update)