标签：std clu 长度 [1] pac har char 参考 tail
题意：给你一个数组a，长度为。有两种操作。一种是改变数组的某个元素的值，一种是满足某种条件的数组b有多少种。条件是：b[i] <= a[i]，并且b[1]^b[2]...^b[n] = k的数组有多少种。数组a的元素都小于1000.
思路：因为数很小，我们把数变成二进制数，然后拆分二进制数。比如1101可以拆成10xx，x可为0可为1，有点像数位dp的试填法。我们对每个a[i]存若干个前缀，记录前缀的长度，以及是这个前缀的二进制数有多少个。然后我们合并相邻的区间，直接暴力二重循环，然后合并。其实这个过程更像dp的过程，感觉只是用了线段树的划分成区间，然后合并的思想。
思路参考这两篇博客：https://blog.csdn.net/jtjy568805874/article/details/56488626， https://blog.csdn.net/qian99/article/details/38171951
代码：
#include <bits/stdc++.h>
#define pii pair<int, int>
#define lowbit(x) (x & (-x))
#define mk make_pair
#define ls(x) (x << 1)
#define rs(x) ((x << 1) | 1)
#define LL long long
using namespace std;
const LL mod = 1000000007;
const int maxn = 20010;
int Log[2010], a[maxn];
struct node {
pii x;
LL tot;
bool operator < (const node& rhs) const {
return x < rhs.x;
}
};
vector<node> tr[maxn * 4], tmp;
int get(int x, int y) {
tr[x].clear();
tr[x].push_back((node){mk(y, 10), 1});
for (int i = y; i; i -= lowbit(i)) {
tr[x].push_back((node){mk(i - lowbit(i), 10 - Log[lowbit(i)]), lowbit(i)});
}
}
void pushup(int x, int l, int r) {
tmp.clear();
tr[x].clear();
for (int i = 0; i < tr[l].size(); i++) {
for (int j = 0; j < tr[r].size(); j++) {
node t1 = tr[l][i], t2 = tr[r][j];
int now = t1.x.first ^ t2.x.first, len = min(t1.x.second, t2.x.second);
now = ((now >> (10 - len)) << (10 - len));
tmp.push_back((node){mk(now, len), (t1.tot * t2.tot) % mod});
}
}
sort(tmp.begin(), tmp.end());
for (int i = 0, j; i < tmp.size(); i = j) {
node tmp1 = (node){tmp[i].x, 0};
for (j = i; j < tmp.size() && tmp[i].x == tmp[j].x; j++) {
tmp1.tot = (tmp1.tot + tmp[j].tot) % mod;
}
tr[x].push_back(tmp1);
}
}
void build(int x, int l, int r) {
if(l == r) {
get(x, a[l]);
return;
}
int mid = (l + r) >> 1;
build(ls(x), l ,mid);
build(rs(x), mid + 1, r);
pushup(x, ls(x), rs(x));
}
LL inv(int x) {
return x == 1 ? 1 : 1ll * inv(mod % x)*(mod - mod / x) % mod;
}
void update(int x, int l, int r, int pos, int val) {
if(l == r) {
get(x, val);
return;
}
int mid = (l + r) >> 1;
if(pos <= mid) update(ls(x), l, mid, pos, val);
else update(rs(x), mid + 1, r, pos, val);
pushup(x, ls(x), rs(x));
}
LL query(int x) {
LL ans = 0;
for (int i = 0; i < tr[1].size(); i++) {
node t = tr[1][i];
if((t.x.first ^ x) >> (10 - t.x.second)) continue;
ans = (ans + (t.tot * inv(1 << (10 - t.x.second))) % mod) % mod;
}
return ans;
}
int main() {
int n, m;
int x, y;
char s[10];
for (int i = 1; i <= 10; i++)
Log[1 << i] = i;
int T;
cin >> T;
while(T--) {
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; i++)
scanf("%d", &a[i]);
build(1, 1, n);
while(m--) {
scanf("%s",s + 1);
if(s[1] == ‘Q‘) {
scanf("%d", &x);
printf("%lld\n", query(x));
} else {
scanf("%d%d", &x, &y);
update(1, 1, n, x + 1, y);
}
}
}
}
标签：std clu 长度 [1] pac har char 参考 tail
原文地址：https://www.cnblogs.com/pkgunboat/p/10534365.html
