[loj2302] noi2017 整数

(30)(bit) 压一起,线段树维护区间左边连续 (0)(2^{30}-1) 的个数。

代码能力太菜了补补水题。

代码

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#include <cstring>
#include <algorithm>
#include <cstdio>

using namespace std;

const int BASE = 1<<30;
const int MX = BASE - 1;
const int MAXN = (30000000+29)/30;

struct {
int len, lc0, lc1, val;
} T[MAXN*4+10];

bool tag0[MAXN*4+10], tag1[MAXN*4+10];
typedef long long ll;

DAT operator+(DAT x, DAT y) {
DAT ret;
ret.len = x.len + y.len;
ret.lc0 = x.lc0; ret.lc1 = x.lc1;
if (x.lc0 == x.len) ret.lc0 = x.len + y.lc0;
if (x.lc1 == x.len) ret.lc1 = x.len + y.lc1;
return ret;
}

void modify0(int rt) {
tag0[rt] = 1;
tag1[rt] = 0;
T[rt].lc1 = T[rt].val = 0;
T[rt].lc0 = T[rt].len;
}

void modify1(int rt) {
tag1[rt] = 1;
tag0[rt] = 0;
T[rt].lc0 = 0;
T[rt].val = MX;
T[rt].lc1 = T[rt].len;
}

void pushDown(int rt) {
if (tag0[rt]) {
modify0(rt<<1);
modify0(rt<<1|1);
tag0[rt] = 0;
}
if (tag1[rt]) {
modify1(rt<<1);
modify1(rt<<1|1);
tag1[rt] = 0;
}
}

void pushUp(int rt) {
T[rt] = T[rt<<1] + T[rt<<1|1];
}

void updc0(int L, int R, int l, int r, int rt) {
if (L > R) return;
if (L <= l && r <= R) {
modify0(rt);
return;
}
pushDown(rt);
int m = (l + r) >> 1;
if (L <= m) updc0(L, R, l, m, rt<<1);
if (R > m) updc0(L, R, m+1, r, rt<<1|1);
pushUp(rt);
}

void updc1(int L, int R, int l, int r, int rt) {
if (L > R) return;
if (L <= l && r <= R) {
modify1(rt);
return;
}
pushDown(rt);
int m = (l + r) >> 1;
if (L <= m) updc1(L, R, l, m, rt<<1);
if (R > m) updc1(L, R, m+1, r, rt<<1|1);
pushUp(rt);
}

void upd(int p, int v, int l, int r, int rt) {
if (l == r) {
if (v == MX) {
T[rt].lc1 = 1;
} else T[rt].lc1 = 0;
if (v == 0) {
T[rt].lc0 = 1;
} else T[rt].lc0 = 0;
T[rt].val = v;
return;
}
pushDown(rt);
int m = (l + r) >> 1;
if (p <= m) upd(p, v, l, m, rt<<1);
else upd(p, v, m+1, r, rt<<1|1);
pushUp(rt);
}

DAT qrys(int L, int R, int l, int r, int rt) {
if (L <= l && r <= R) {
return T[rt];
}
pushDown(rt);
int m = (l + r) >> 1;
DAT ret; ret.len = ret.lc0 = ret.lc1 = 0;
if (L <= m) ret = ret + qrys(L, R, l, m, rt<<1);
if (R > m) ret = ret + qrys(L, R, m+1, r, rt<<1|1);
return ret;
}

int qry(int p, int l, int r, int rt) {
if (l == r) return T[rt].val;
int m = (l + r) >> 1;
pushDown(rt);
if (p <= m) return qry(p, l, m, rt<<1);
else return qry(p, m+1, r, rt<<1|1);
}

void add(int p, int v) {
if (!v) return;
int t = qry(p, 0, MAXN, 1);
if ((t+v) >= BASE) {
upd(p, (t+v)%BASE, 0, MAXN, 1);
DAT dat = qrys(p+1, MAXN, 0, MAXN, 1);
updc0(p+1, p+dat.lc1, 0, MAXN, 1);
upd(p+1+dat.lc1, qry(p+1+dat.lc1, 0, MAXN, 1) + 1, 0, MAXN, 1);
} else upd(p, t+v, 0, MAXN, 1);
}

void sub(int p, int v) {
if (!v) return;
int t = qry(p, 0, MAXN, 1);
if ((t-v) < 0) {
upd(p, t-v+BASE, 0, MAXN, 1);
DAT dat = qrys(p+1, MAXN, 0, MAXN, 1);
updc1(p+1, p+dat.lc0, 0, MAXN, 1);
upd(p+1+dat.lc0, qry(p+1+dat.lc0, 0, MAXN, 1) - 1, 0, MAXN, 1);
} else upd(p, t-v, 0, MAXN, 1);
}

void build(int l, int r, int rt) {
if (l == r) {
T[rt].len = T[rt].lc0 = 1;
return;
}
int m = (l + r) >> 1;
build(l, m, rt<<1);
build(m+1, r, rt<<1|1);
pushUp(rt);
}

int n, t1, t2, t3;

int main() {
scanf("%d%d%d%d", &n, &t1, &t2, &t3);
build(0, MAXN, 1);
for (int i = 1; i <= n; i++) {
int opt;
scanf("%d", &opt);
if (opt == 1) {
int a, b;
scanf("%d%d", &a, &b);

//2^{b%30}*b
if (a < 0) {
a = -a;
ll v = (1ll<<(b%30))*a;
sub(b/30, v%BASE);
sub(b/30+1, v/BASE);
} else {
ll v = (1ll<<(b%30))*a;
add(b/30, v%BASE);
add(b/30+1, v/BASE);
}
} else {
int k;
scanf("%d", &k);
int t = qry(k/30, 0, MAXN, 1);
printf("%dn", (t>>(k%30))&1);
}
}
return 0;
}