A. water the garden
Code
#include <bits/stdc++.h>
#define maxn 210
using namespace std;
typedef long long LL;
int n, k;
bool vis[maxn];
int x[maxn];
bool ok() {
for (int i = 1; i <= n; ++i) if (!vis[i]) return false;
return true;
}
void work() {
scanf("%d%d", &n,&k);
memset(vis, 0, sizeof vis);
for (int i = 0; i < k; ++i) scanf("%d", &x[i]);
int cnt = 0;
while (true) {
if (ok()) { printf("%d\n", cnt); return; }
for (int i = 0; i < k; ++i) {
int p1 = max(0, x[i]-cnt),
p2 = min(n, x[i]+cnt);
vis[p1] = vis[p2] = true;
}
++cnt;
}
}
int main() {
int T;
scanf("%d", &T);
while (T--) work();
return 0;
}B. Tea Queue
题意
若干个人排队等茶喝，每个人有到达时间和离去时间，每个时刻只能一个人喝茶。
问每个人喝到茶的时间。
思路
排序，然后按顺序模拟（将当前时间逐个向后推移）
Code
#include <bits/stdc++.h>
#define maxn 1010
using namespace std;
typedef long long LL;
struct node {
int l, r, id;
}a[maxn];
int ans[maxn];
bool cmp(node a, node b) {
return (a.l < b.l) || (a.l==b.l && a.id<b.id);
}
void work() {
int n;
scanf("%d", &n);
for (int i = 0; i < n; ++i) {
scanf("%d%d", &a[i].l, &a[i].r);
a[i].id = i;
}
sort(a, a+n, cmp);
int time = 0;
for (int i = 0; i < n; ++i) {
if (time <= a[i].l) {
time = a[i].l;
ans[a[i].id] = time;
++time;
}
else {
if (time <= a[i].r) {
ans[a[i].id] = time;
++time;
}
else {
ans[a[i].id] = 0;
}
}
}
printf("%d", ans[0]);
for (int i = 1; i < n; ++i) printf(" %d", ans[i]); puts("");
}
int main() {
int T;
scanf("%d", &T);
while (T--) work();
return 0;
}
C. Swap Adjacent Elements
题意
给出 1-n 的一个排列，只能交换相邻两个元素，即 \(i\) 与 \(i+1\)，且只有给定的位置 \(i\) 才可交换。
问能否通过交换得到一个升序序列。
思路
假设元素 \(a[i]\) 现在处在 \(i\) 位置，那么交换必然通过 \(i-a[i]\) 的一整段线段，也即要求这一整段上的点都是可交换的位置。
即转化成一个 线段覆盖 问题，用线段树解决。
Code
#include <bits/stdc++.h>
#define maxn 200010
#define lson (rt<<1)
#define rson (lson|1)
using namespace std;
typedef long long LL;
struct node { int l, r; bool flag; }tr[maxn<<2];
char s[maxn];
int a[maxn];
void build(int rt, int l, int r) {
tr[rt].l = l, tr[rt].r = r;
if (l == r) {
tr[rt].flag = s[l] == '1'; return;
}
int mid = l+r>>1;
build(lson, l, mid); build(rson, mid+1, r);
tr[rt].flag = tr[lson].flag & tr[rson].flag;
}
bool query(int rt, int l, int r) {
if (tr[rt].l==l&&tr[rt].r==r) return tr[rt].flag;
int mid = tr[rt].l+tr[rt].r >> 1;
if (r <= mid) return query(lson, l, r);
else if (l > mid) return query(rson, l, r);
else return query(lson, l, mid) & query(rson, mid+1, r);
}
int main() {
int n;
scanf("%d", &n);
for (int i = 1; i <= n; ++i) {
scanf("%d", &a[i]);
}
scanf("%s", s+1);
build(1, 1, n);
for (int i = 1; i <= n; ++i) {
if (i == a[i]) continue;
if (i < a[i] && !query(1, i, a[i]-1)) { puts("NO"); return 0; }
else if (i > a[i] && !query(1, a[i], i-1)) { puts("NO"); return 0; }
}
puts("YES");
return 0;
}
F. SUM and REPLACE
题意
对一个序列进行两种操作：
- 将 \([l,r]\) 中每个数 \(x\) 变为其约数个数 \(D(x)\)
- 对 \([l,r]\) 区间求和
思路
神似 bzoj 3211 花神游历各国
Code
Ver. 1：线段树
#include <bits/stdc++.h>
#define maxn 300010
#define maxl 1000010
#define lson (rt << 1)
#define rson (rt << 1 | 1)
using namespace std;
typedef long long LL;
int prime[maxn], d[maxl], cnt[maxl], n, k, a[maxn];
bool check[maxl];
void init() {
int tot = 0; d[1] = 1;
for (int i = 2; i <= 1000000; ++i) {
if (!check[i]) {
prime[tot++] = i;
d[i] = 2, cnt[i] = 1;
}
for (int j = 0; j < tot; ++j) {
if (i * prime[j] > 1000000) break;
check[i * prime[j]] = true;
if (i % prime[j] == 0) {
cnt[i * prime[j]] = cnt[i] + 1;
d[i * prime[j]] = d[i] / (cnt[i] + 1) * (cnt[i * prime[j]] + 1);
break;
}
cnt[i * prime[j]] = 1;
d[i * prime[j]] = d[i] << 1;
}
}
}
struct node { int l, r; bool flag; LL sum; } tr[maxn<<2];
inline void push_up(int rt) {
tr[rt].sum = tr[lson].sum + tr[rson].sum;
tr[rt].flag = tr[lson].flag & tr[rson].flag;
}
inline int midi(int a, int b) { return a + b >> 1; }
void build(int rt, int l, int r) {
tr[rt].l = l, tr[rt].r = r, tr[rt].flag = 0;
if (l == r) {
scanf("%I64d", &tr[rt].sum);
if (tr[rt].sum <= 2) tr[rt].flag = 1;
return;
}
int mid = midi(l,r);
build(lson, l, mid); build(rson, mid + 1, r);
push_up(rt);
}
void modify(int rt, int l, int r) {
if (tr[rt].flag) return;
if (tr[rt].l == tr[rt].r) {
tr[rt].sum = d[tr[rt].sum];
if (tr[rt].sum <= 2) tr[rt].flag = 1;
return;
}
int mid = midi(tr[rt].l, tr[rt].r);
if (r <= mid) modify(lson, l, r);
else if (l > mid) modify(rson, l, r);
else { modify(lson, l, mid); modify(rson, mid + 1, r); }
push_up(rt);
}
LL query(int rt, int l, int r) {
if (tr[rt].l == l && tr[rt].r == r) return tr[rt].sum;
int mid = midi(tr[rt].l, tr[rt].r);
if (r <= mid) return query(lson, l, r);
else if (l > mid) return query(rson, l, r);
else return query(lson, l, mid) + query(rson, mid + 1, r);
}
int main() {
scanf("%d%d", &n,&k);
build(1, 1, n);
init();
while (k--) {
int t, l, r;
scanf("%d%d%d", &t,&l,&r);
if (t==2) printf("%I64d\n", query(1, l, r));
else modify(1, l, r);
}
return 0;
}
Ver. 2：树状数组+并查集
#include <bits/stdc++.h>
#define maxn 300010
#define maxl 1000010
using namespace std;
typedef long long LL;
int prime[maxn], d[maxl], cnt[maxl], fa[maxn], n, k, a[maxn];
bool check[maxl];
LL c[maxn];
void init() {
int tot = 0; d[1] = 1;
for (int i = 2; i <= 1000000; ++i) {
if (!check[i]) {
prime[tot++] = i;
d[i] = 2, cnt[i] = 1;
}
for (int j = 0; j < tot; ++j) {
if (i * prime[j] > 1000000) break;
check[i * prime[j]] = true;
if (i % prime[j] == 0) {
cnt[i * prime[j]] = cnt[i] + 1;
d[i * prime[j]] = d[i] / (cnt[i] + 1) * (cnt[i * prime[j]] + 1);
break;
}
cnt[i * prime[j]] = 1;
d[i * prime[j]] = d[i] << 1;
}
}
}
int lowbit(int x) { return x & (-x); }
void add(int p, int x) {
while (p <= n) c[p] += x, p += lowbit(p);
}
LL query(int x) {
LL ret = 0;
while (x) ret += c[x], x -= lowbit(x);
return ret;
}
int find(int x) { return fa[x] == x ? x : fa[x] = find(fa[x]); }
int main() {
init();
scanf("%d%d", &n,&k);
for (int i = 1; i <= n; ++i) {
scanf("%d", &a[i]);
add(i, a[i]);
fa[i] = i;
}
fa[n+1] = n+1;
for (int i = 1; i <= n; ++i) if (a[i] <= 2) fa[i] = find(i+1);
while (k--) {
int t, l, r;
scanf("%d%d%d", &t,&l,&r);
if (t==2) printf("%I64d\n", query(r)-query(l-1));
else for (int i = find(l); i <= r; i = find(i+1)) {
add(i, d[a[i]] - a[i]);
if ((a[i] = d[a[i]]) <= 2) fa[i] = find(i+1);
}
}
return 0;
}
