题目链接:codeforces242E. XOR on Segment
题意:
给定一个序列,然后有两种操作:
- 区间求和。
- 区间每个数异或x。
题解:
把每个数写成二进制形式,然后按二进制位建线段树,即建20棵线段树。然后对于每棵线段树维护区间和,以及区间异或。因为拆分之后,每一位只有可能是0或1,所以可以直接维护区间翻转即可。
参考代码:
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAXN = 1e5 + 10;
int n, m;
int a[MAXN];
int seg[MAXN << 2][25], lazy[MAXN << 2][25];
void pushup(int rt)
{
for (int i = 0; i <= 20; i++)
seg[rt][i] = seg[rt << 1][i] + seg[rt << 1 | 1][i];
}
void pushdown(int l, int r, int rt)
{
for (int i = 0; i <= 20; i++)
if (lazy[rt][i])
{
lazy[rt << 1][i] ^= lazy[rt][i];
lazy[rt << 1 | 1][i] ^= lazy[rt][i];
int mid = (l + r) >> 1;
seg[rt << 1][i] = (mid - l + 1) - seg[rt << 1][i];
seg[rt << 1 | 1][i] = (r - mid) - seg[rt << 1 | 1][i];
lazy[rt][i] = 0;
}
}
void build(int l, int r, int rt)
{
if (l == r)
{
for (int i = 0; i <= 20; i++)
{
seg[rt][i] = (a[l] >> i) & 1;
lazy[rt][i] = 0;
}
return;
}
int mid = (l + r) >> 1;
build(l, mid, rt << 1);
build(mid + 1, r, rt << 1 | 1);
pushup(rt);
}
void update(int L, int R, int pos, int l, int r, int rt)
{
if (L <= l && r <= R)
{
lazy[rt][pos] ^= 1;
seg[rt][pos] = (r - l + 1) - seg[rt][pos];
return;
}
pushdown(l, r, rt);
int mid = (l + r) >> 1;
if (L <= mid)
update(L, R, pos, l, mid, rt << 1);
if (R > mid)
update(L, R, pos, mid + 1, r, rt << 1 | 1);
pushup(rt);
}
ll query(int L, int R, int pos, int l, int r, int rt)
{
if (L <= l && r <= R)
return seg[rt][pos];
pushdown(l, r, rt);
int mid = (l + r) >> 1;
ll res = 0;
if (L <= mid)
res += query(L, R, pos, l, mid, rt << 1);
if (R > mid)
res += query(L, R, pos, mid + 1, r, rt << 1 | 1);
return res;
}
int main()
{
scanf("%d", &n);
for (int i = 1; i <= n; i++)
scanf("%d", &a[i]);
build(1, n, 1);
scanf("%d", &m);
while (m--)
{
int op, l, r;
scanf("%d%d%d", &op, &l, &r);
if (op == 1)
{
ll ans = 0;
for (int i = 0; i <= 20; i++)
ans += query(l, r, i, 1, n, 1) * (1ll << i);
printf("%lld\n", ans);
}
else
{
int x;
scanf("%d", &x);
for (int i = 0; i <= 20; i++)
if ((x >> i) & 1)
update(l, r, i, 1, n, 1);
}
}
return 0;
}