HDU

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题意:给定n 个数和 q 个查询， qi 为查询数，求去掉下标为qi 的元素后其他元素 and , or , xor的结果

分析:对于xor 直接 ^ a[qi] 即可，其它则考虑a[qi] 二进制每一位为1或0的情况，当只有这个数当前位为1 的时候，or 结果去掉当前位，当只有这个数当前位为0的时候，and 加上当前位

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <map>using namespace std;typedef long long int ll;map<int,int>mp;#define maxn 100005int n , q;ll c[maxn];ll _and,_or,_xor;void deal(ll a){    int cnt = 0;    while( a )    {        int x = a%2 ;        a/=2;        if( x )mp[cnt]++;        cnt++;    }}void dealAnd(ll a){   for(int i = 0 ; i < 30 ; ++i)   {      if( mp[i] == n - 1 && (a & (1<<i)) == 0 )      {          _and|=(1<<i);      }   }}void dealOr(int a){   for(int i = 0 ; i < 30 ; ++i)   {      if( mp[i] == 1 && (a & (1<<i)) )      {          _or^=(1<<i);      }   }}int main(){        while( cin >> n >> q )    {        mp.clear();        _and = _or = _xor = 0;        for(int i = 1 ; i <= n ; ++i)        {            scanf("%lld",c+i);            _xor ^= c[i];            _and &= c[i];            _or |= c[i];            if( i == 1 )                _and = _or = _xor = c[i];            deal(c[i]);        }        ll t1,t2,t3;        t1 = _xor;        t2 = _or;        t3 = _and;        for(int i = 0 ; i < q ; ++i)        {            int a;            _xor = t1;            _or = t2;            _and = t3;            scanf("%d",&a);            if( n == 1 )            {                printf("0 0 0\n");                continue;            }            _xor ^= c[a];            dealAnd(c[a]);            dealOr(c[a]);            cout << _and << " " << _or << " " << _xor << endl;        }    }}

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