HDU 6186 CS Course（前缀+后缀）

http://acm.hdu.edu.cn/showproblem.php?pid=6186

题意：
给出n个数，共有n次询问，每次询问给出一个数p，求除去第p个数后的n-1个数的&、|、^值。

思路：
分别计算出&、|、^的前缀和后缀，将前缀和后缀相计算即可。

 #include<iostream>
 #include<cstdio>
 #include<algorithm>
 using namespace std;
 const int maxn = 1e5+;

 int n, q;
 int a[maxn], prea[maxn], suffa[maxn], preo[maxn], suffo[maxn], prex[maxn], suffx[maxn];

 int main()
 {
     //freopen("in.txt","r",stdin);
     while(~scanf("%d%d",&n,&q))
     {
         scanf("%d",&a[]);
         prea[] = preo[] = prex[] = a[];
         for(int i=;i<=n;i++)
         {
             scanf("%d",&a[i]);
             prea[i] = a[i] & prea[i-];
             preo[i] = a[i] | preo[i-];
             prex[i] = a[i] ^ prex[i-];
         }
         suffa[n] = suffo[n] = suffx[n] = a[n];
         for(int i=n-;i>=;i--)
         {
             suffa[i] = a[i] & suffa[i+];
             suffo[i] = a[i] | suffo[i+];
             suffx[i] = a[i] ^ suffx[i+];
         }
         while(q--)
         {
             int pos; scanf("%d",&pos);
             int ans1 = prea[pos-] & suffa[pos+];
             int ans2 = preo[pos-] | suffo[pos+];
             int ans3 = prex[pos-] ^ suffx[pos+];
             if(pos==)
             {
                 ans1 = suffa[pos+];
                 ans2 = suffo[pos+];
                 ans3 = suffx[pos+];
             }
             if(pos==n)
             {
                 ans1 = prea[pos-];
                 ans2 = preo[pos-];
                 ans3 = prex[pos-];
             }
             printf("%d %d %d\n", ans1, ans2, ans3);
         }
     }
     return ;
 }