标签：AC The xor 有一个 art answer class name xor sum
Time limit : 2sec / Memory limit : 1024MB
Score : 500 points
There is an integer sequence A of length N.
Find the number of the pairs of integers l and r (1≤l≤r≤N) that satisfy the following condition:
Here, xor denotes the bitwise exclusive OR.
Definition of XORInput is given from Standard Input in the following format:
N A1 A2 … AN
Print the number of the pairs of integers l and r (1≤l≤r≤N) that satisfy the condition.
4 2 5 4 6
5
(l,r)=(1,1),(2,2),(3,3),(4,4) clearly satisfy the condition. (l,r)=(1,2) also satisfies the condition, since A1 xor A2=A1 + A2=7. There are no other pairs that satisfy the condition, so the answer is 5.
9 0 0 0 0 0 0 0 0 0
45
19 885 8 1 128 83 32 256 206 639 16 4 128 689 32 8 64 885 969 1
37
转化成位运算，每位只能有一个
#include <iostream> #include <algorithm> #include <cstring> #include <cstdio> #include <vector> #include <queue> #include <stack> #include <cstdlib> #include <iomanip> #include <cmath> #include <cassert> #include <ctime> #include <map> #include <set> using namespace std; #pragma comment(linker, "/stck:1024000000,1024000000") #define lowbit(x) (x&(-x)) #define max(x,y) (x>=y?x:y) #define min(x,y) (x<=y?x:y) #define MAX 100000000000000000 #define MOD 1000000007 #define pi acos(-1.0) #define ei exp(1) #define PI 3.1415926535897932384626433832 #define ios() ios::sync_with_stdio(true) #define INF 0x3f3f3f3f #define mem(a) (memset(a,0,sizeof(a))) typedef long long ll; int ans[26],n,a[200006]; ll pos=0,cnt=0; int main() { scanf("%d",&n); int k=1; for(int i=1;i<=n;i++) { scanf("%d",&a[i]); int flag=1; for(int j=0;j<=20;j++) { if(a[i]&(1<<j)) ans[j]++; if(ans[j]>1) {flag=0;} } if(!flag) { pos+=cnt; while(k<i) { int ok=1; for(int j=0;j<=20;j++) { ans[j]-=(a[k]&(1<<j))?1:0; if(ans[j]>1) ok=0; } k++; cnt--; if(ok) break; else pos+=cnt; } } cnt++; } printf("%lld\n",pos+((1+cnt)*cnt/2)); return 0; }
标签：AC The xor 有一个 art answer class name xor sum
原文地址：https://www.cnblogs.com/shinianhuanniyijuhaojiubujian/p/9094400.html
