FZU2105 Digits Count(按位建线段树)题解
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题意:
给出区间与、或、异或\(x\)操作,还有询问区间和。
思路:
因为数比较小,我们给每一位建线段树,这样每次只要更新对应位的答案。
与\(0\)和或\(1\)相当于重置区间,异或\(1\)相当于翻转区间,那么设出两个\(lazy\)搞一下。注意父区间\(pushdown\)重置标记时,子区间的翻转标记要清空。
代码:
#include <map>
#include <set>
#include <queue>
#include <cmath>
#include <stack>
#include <ctime>
#include <vector>
#include <cstdio>
#include <string>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
const int maxn = 1e6 + 5;
const int MAXM = 3e6;
const ll MOD = 1e9 + 7;
const ull seed = 131;
const int INF = 0x3f3f3f3f;
#define lson (rt << 1)
#define rson (rt << 1 | 1)
int sum[5][maxn << 2];
int lazy[5][maxn << 2], rev[5][maxn << 2];
int a[maxn];
void pushup(int rt, int bit)
sum[bit][rt] = sum[bit][lson] + sum[bit][rson];
void pushdown(int rt, int bit, int l, int r)
int m = (l + r) >> 1;
if(lazy[bit][rt] != -1)
sum[bit][lson] = lazy[bit][rt] * (m - l + 1);
sum[bit][rson] = lazy[bit][rt] * (r - m);
lazy[bit][lson] = lazy[bit][rson] = lazy[bit][rt];
rev[bit][lson] = rev[bit][rson] = 0; //!!!!!
lazy[bit][rt] = -1;
if(rev[bit][rt])
sum[bit][lson] = m - l + 1 - sum[bit][lson];
sum[bit][rson] = r - m - sum[bit][rson];
rev[bit][lson] ^= 1;
rev[bit][rson] ^= 1;
rev[bit][rt] = 0;
void build(int l, int r, int bit, int rt)
lazy[bit][rt] = -1;
rev[bit][rt] = 0;
if(l == r)
sum[bit][rt] = (a[l] >> bit) & 1;
return;
int m = (l + r) >> 1;
build(l, m, bit, lson);
build(m + 1, r, bit, rson);
pushup(rt, bit);
void update(int L, int R, int l, int r, int op, int bit, int rt)
if(L <= l && R >= r)
if(op == 1) //&0
lazy[bit][rt] = 0;
rev[bit][rt] = 0;
sum[bit][rt] = 0;
else if(op == 2) //|1
lazy[bit][rt] = 1;
rev[bit][rt] = 0;
sum[bit][rt] = r - l + 1;
else //^1
rev[bit][rt] ^= 1;
sum[bit][rt] = r - l + 1 - sum[bit][rt];
return;
pushdown(rt, bit, l, r);
int m = (l + r) >> 1;
if(L <= m)
update(L, R, l, m, op, bit, lson);
if(R > m)
update(L, R, m + 1, r, op, bit, rson);
pushup(rt, bit);
int query(int L, int R, int l, int r, int bit, int rt)
if(L <= l && R >= r)
return sum[bit][rt];
pushdown(rt, bit, l, r);
int m = (l + r) >> 1, ret = 0;
if(L <= m)
ret += query(L, R, l, m, bit, lson);
if(R > m)
ret += query(L, R, m + 1, r, bit, rson);
return ret;
int main()
int n, m, T;
scanf("%d", &T);
while(T--)
scanf("%d%d", &n, &m);
for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
for(int i = 0; i < 5; i++) build(1, n, i, 1);
while(m--)
int op, l, r, x;
char ope[10];
scanf("%s", &ope);
if(ope[0] == 'A') op = 1;
else if(ope[0] == 'O') op = 2;
else if(ope[0] == 'X') op = 3;
else op = 4;
if(op == 4)
scanf("%d%d", &l, &r);
l++, r++;
ll ans = 0;
for(int i = 0; i < 5; i++)
x = query(l, r, 1, n, i, 1);
ans += 1LL * (1LL << i) * x;
printf("%lld\n", ans);
else
scanf("%d%d%d", &x, &l, &r);
l++, r++;
if(op == 2 || op == 3)
for(int i = 0; i < 5; i++)
if((x >> i) & 1)
update(l, r, 1, n, op, i, 1);
else
for(int i = 0; i < 5; i++)
if(!((x >> i) & 1))
update(l, r, 1, n, op, i, 1);
return 0;
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