CS Course
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 338 Accepted Submission(s): 167
Problem Description
Little A has come to college and majored in Computer and Science.
Today he has learned bit-operations in Algorithm Lessons, and he got a problem as homework.
Here is the problem:
You are giving n non-negative integers a1,a2,⋯,an, and some queries.
A query only contains a positive integer p, which means you
are asked to answer the result of bit-operations (and, or, xor) of all the integers except ap.
Today he has learned bit-operations in Algorithm Lessons, and he got a problem as homework.
Here is the problem:
You are giving n non-negative integers a1,a2,⋯,an, and some queries.
A query only contains a positive integer p, which means you
are asked to answer the result of bit-operations (and, or, xor) of all the integers except ap.
Input
There are no more than 15 test cases.
Each test case begins with two positive integers n and p
in a line, indicate the number of positive integers and the number of queries.
2≤n,q≤105
Then n non-negative integers a1,a2,⋯,an follows in a line, 0≤ai≤109 for each i in range[1,n].
After that there are q positive integers p1,p2,⋯,pqin q lines, 1≤pi≤n for each i in range[1,q].
Each test case begins with two positive integers n and p
in a line, indicate the number of positive integers and the number of queries.
2≤n,q≤105
Then n non-negative integers a1,a2,⋯,an follows in a line, 0≤ai≤109 for each i in range[1,n].
After that there are q positive integers p1,p2,⋯,pqin q lines, 1≤pi≤n for each i in range[1,q].
Output
For
each query p, output three non-negative integers indicates the result
of bit-operations(and, or, xor) of all non-negative integers except ap in a line.
Sample Input
3 3
1 1 1
1
2
3
Sample Output
1 1 0
1 1 0
1 1 0
求除给定数之外剩下的数的& | 异或值
#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; #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 3.141592653589793238462 #define ios() ios::sync_with_stdio(false) #define INF 1044266558 #define mem(a) (memset(a,0,sizeof(a))) typedef long long ll; const int M=1134217727; int n,a[1000005],ans,ano,anx; int sans,sano,sanx,q,pos[50],x,len; int main() { while(scanf("%d%d",&n,&q)!=EOF) { ano=0,anx=0,ans=M; memset(pos,0,sizeof(pos)); for(int i=1;i<=n;i++) { scanf("%d",&a[i]); x=a[i]; ans&=a[i]; ano|=a[i]; anx^=a[i]; len=0; while(x) { pos[len++]+=x%2; x>>=1; } } while(q--) { scanf("%d",&x); sans=ans;sano=ano;sanx=anx; x=a[x]; sanx^=x; for(int i=0;i<=30;i++) { if(pos[i]==n-1 && !(x%2)) sans+=(1<<i); if(pos[i]==1 && x%2) sano-=(1<<i); x>>=1; } printf("%d %d %d ",sans,sano,sanx); } } return 0; }