codechef Chef And Easy Xor Queries

做法：我们考虑前缀异或和，修改操作就变成了区间[i,n]都异或x

查询操作就变成了:区间[1,x]中有几个k

显然的分块，每个块打一个tag标记表示这个块中所有的元素都异或了tag[x]

然后处理出这个块中每种数的个数

注意查询的时候零散的块要下放标记

代码：

#include<bits/stdc++.h>
#define N 100005
#define M 334
using namespace std;
int cnt[M][N*14],sum[N],tag[M],ll[M],rr[M],a[N];
int n,Q,opt,x,y,block,num;
inline int gt(int x){return (x-1)/block+1;}
inline void pushdown(int x){
	if (!tag[x]) return;
	for (int i=ll[x];i<=rr[x];i++){
		cnt[x][sum[i]]--;
		sum[i]^=tag[x];
		cnt[x][sum[i]]++;
	}
	tag[x]=0;
}
inline void query(int x,int y,int k){
	//printf("%d %d\n",gt(x),gt(y));
	int ans=0;
	pushdown(gt(x));pushdown(gt(y));
	if (gt(x)+1>=gt(y)){
		for (int i=x;i<=y;i++) if (sum[i]==k) ans++;
	}
	else {
		for (int i=x;i<=rr[gt(x)];i++) if (sum[i]==k) ans++;
		for (int i=ll[gt(y)];i<=y;i++) if (sum[i]==k) ans++;
		for (int j=gt(x)+1;j<gt(y);j++) ans+=cnt[j][k^tag[j]];
	}
	printf("%d\n",ans);
}
inline void change(int x,int y,int z){
	int tmp=z;z=z^a[x];a[x]=tmp;
	if (gt(x)+1>=gt(y)){
		for (int i=x;i<=y;i++){
			cnt[gt(i)][sum[i]]--;
			sum[i]^=z;
			cnt[gt(i)][sum[i]]++;
		}
	}
	else {
		for (int i=x;i<=rr[gt(x)];i++){
			cnt[gt(i)][sum[i]]--;
			sum[i]^=z;
			cnt[gt(i)][sum[i]]++;
		}
		for (int i=ll[gt(y)];i<=y;i++){
			cnt[gt(i)][sum[i]]--;
			sum[i]^=z;
			cnt[gt(i)][sum[i]]++;
		}
		for (int j=gt(x)+1;j<gt(y);j++) tag[j]^=z;
	}
	//for (int i=1;i<=n;i++) printf("%d %d\n",sum[i],tag[i]);
}
int main(){
	scanf("%d%d",&n,&Q);
	for (int i=1;i<=n;i++) scanf("%d",&a[i]),sum[i]=sum[i-1]^a[i];
	block=(int)sqrt(n);num=n/block;if (n%block) num++;
	for (int i=1;i<=num;i++) ll[i]=(i-1)*block+1,rr[i]=i*block;rr[num]=n;
	//for (int i=1;i<=num;i++) printf("%d %d\n",ll[i],rr[i]);
	for (int i=1;i<=num;i++)
		for (int j=ll[i];j<=rr[i];j++) cnt[i][sum[j]]++;
	while (Q--){
		scanf("%d%d%d",&opt,&x,&y);
		if (opt==2) query(1,x,y);
		else change(x,n,y);
	}
	return 0;
} 

　　

1. #include<bits/stdc++.h>
2. #define N 100005
3. #define M 334
4. using namespace std;
5. int cnt[M][N*14],sum[N],tag[M],ll[M],rr[M],a[N];
6. int n,Q,opt,x,y,block,num;
7. inline int gt(int x){return (x-1)/block+1;}
8. inline void pushdown(int x){
9. if (!tag[x]) return;
10. for (int i=ll[x];i<=rr[x];i++){
11. cnt[x][sum[i]]--;
12. sum[i]^=tag[x];
13. cnt[x][sum[i]]++;
14. }
15. tag[x]=0;
16. }
17. inline void query(int x,int y,int k){
18. //printf("%d %d\n",gt(x),gt(y));
19. int ans=0;
20. pushdown(gt(x));pushdown(gt(y));
21. if (gt(x)+1>=gt(y)){
22. for (int i=x;i<=y;i++) if (sum[i]==k) ans++;
23. }
24. else {
25. for (int i=x;i<=rr[gt(x)];i++) if (sum[i]==k) ans++;
26. for (int i=ll[gt(y)];i<=y;i++) if (sum[i]==k) ans++;
27. for (int j=gt(x)+1;j<gt(y);j++) ans+=cnt[j][k^tag[j]];
28. }
29. printf("%d\n",ans);
30. }
31. inline void change(int x,int y,int z){//[x,y]����^z
32. int tmp=z;z=z^a[x];a[x]=tmp;
33. if (gt(x)+1>=gt(y)){
34. for (int i=x;i<=y;i++){
35. cnt[gt(i)][sum[i]]--;
36. sum[i]^=z;
37. cnt[gt(i)][sum[i]]++;
38. }
39. }
40. else {
41. for (int i=x;i<=rr[gt(x)];i++){
42. cnt[gt(i)][sum[i]]--;
43. sum[i]^=z;
44. cnt[gt(i)][sum[i]]++;
45. }
46. for (int i=ll[gt(y)];i<=y;i++){
47. cnt[gt(i)][sum[i]]--;
48. sum[i]^=z;
49. cnt[gt(i)][sum[i]]++;
50. }
51. for (int j=gt(x)+1;j<gt(y);j++) tag[j]^=z;
52. }
53. //for (int i=1;i<=n;i++) printf("%d %d\n",sum[i],tag[i]);
54. }
55. int main(){
56. scanf("%d%d",&n,&Q);
57. for (int i=1;i<=n;i++) scanf("%d",&a[i]),sum[i]=sum[i-1]^a[i];
58. block=(int)sqrt(n);num=n/block;if (n%block) num++;
59. for (int i=1;i<=num;i++) ll[i]=(i-1)*block+1,rr[i]=i*block;rr[num]=n;
60. //for (int i=1;i<=num;i++) printf("%d %d\n",ll[i],rr[i]);
61. for (int i=1;i<=num;i++)
62. for (int j=ll[i];j<=rr[i];j++) cnt[i][sum[j]]++;
63. while (Q--){
64. scanf("%d%d%d",&opt,&x,&y);
65. if (opt==2) query(1,x,y);
66. else change(x,n,y);
67. }
68. return 0;
69. }

来源：https://www.cnblogs.com/ckr1225/p/9010059.html