題干：

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,?,ana1,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?apap.?

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≤1052≤n,q≤105?

Then n non-negative integers?a1,a2,?,ana1,a2,?,an?follows in a line,?0≤ai≤1090≤ai≤109?for each i in range[1,n].?

After that there are q positive integers?p1,p2,?,pqp1,p2,?,pqin q lines,?1≤pi≤n1≤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?apap?in a line.?

Sample Input

3 3 1 1 1 1 2 3

Sample Output

1 1 0 1 1 0 1 1 0

題目大意：

題意:給定n 個數和 q 個查詢，qi 為查詢數，求去掉下標為qi 的元素后其他元素 and , or , xor的結果。

解題報告：

這題也可以用線段樹nlogn搞或者直接處理前綴后綴然后on搞（因為沒有修改嘛）然后也可以按位統計然后nlogn亂搞。。

但是前者沒法做帶更新的，后者可以。

AC代碼：

#include<cstdio> #include<iostream> #include<algorithm> #include<queue> #include<map> #include<vector> #include<set> #include<string> #include<cmath> #include<cstring> #define F first #define S second #define ll long long #define pb push_back #define pm make_pair using namespace std; typedef pair<int,int> PII; const int MAX = 2e5 + 5; int cnt[44]; ll a[MAX]; int n,q,p; int main() {while(~scanf("%d%d",&n,&q)) {memset(cnt,0,sizeof cnt);ll Xor = 0;for(int i = 1; i<=n; i++) scanf("%lld",a+i),Xor ^= a[i];for(int i = 1; i<=n; i++) {for(int j = 0; j<33; j++) {cnt[j] += (a[i]>>j)&1;}}while(q--) {scanf("%d",&p);ll And=0,Or=0,X;ll tar = a[p];X = Xor^tar;for(int j = 0; j<33; j++) {ll tmp = (tar>>j)&1;if(cnt[j]-tmp>0) Or |= (1<<j);if(cnt[j]-tmp == n-1) And |= (1<<j);}printf("%lld %lld %lld\n",And,Or,X);} }return 0 ; }

?

AC代碼2：

#include<iostream> #include<cstdio> #include<cstring> #define ll long long #define mid ((st[id].l+st[id].r)>>1) #define lson (id<<1) #define rson ((id<<1)|1)using namespace std;const int N = 150000; int arr[N]; struct Node{int l, r, OR, XOR, AND; }st[N<<3];void build(int id, int l, int r) {st[id].l = l; st[id].r = r;if(l == r){st[id].OR = arr[l];st[id].AND = arr[l];st[id].XOR = arr[l];return;}build(lson, l, mid);build(rson, mid+1, r);st[id].OR = st[lson].OR | st[rson].OR;st[id].XOR = st[lson].XOR ^ st[rson].XOR;st[id].AND = st[lson].AND & st[rson].AND; }int query(int id, int l, int r, int op){if(st[id].l == l && st[id].r == r){if(op == 1)return st[id].OR;if(op == 2)return st[id].XOR;if(op == 3)return st[id].AND;}if(l > mid)return query(rson, l, r, op);else if(r <= mid)return query(lson, l, r, op);else{if(op == 1)return query(lson, l, mid, op) | query(rson, mid+1, r, op);if(op == 2)return query(lson, l, mid, op) ^ query(rson, mid+1, r, op);if(op == 3)return query(lson, l, mid, op) & query(rson, mid+1, r, op);} }int main() {int n, q;while(scanf("%d%d", &n, &q)!=EOF){for(int i = 1; i <= n; i++)scanf("%d", &arr[i]);build(1, 1, n);int p;while(q--){scanf("%d", &p);if(p == 1)printf("%d %d %d\n", query(1, 2, n, 3), query(1, 2, n, 1), query(1, 2, n, 2));else if(p == n)printf("%d %d %d\n", query(1, 1, n-1, 3), query(1, 1, n-1, 1), query(1, 1, n-1, 2));else printf("%d %d %d\n", query(1, 1, p-1, 3)&query(1, p+1, n, 3), query(1, 1, p-1, 1)|query(1, p+1, n, 1), query(1, 1, p-1, 2)^query(1, p+1, n, 2));}}return 0; }

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