开局有人过C，我就去钻研C了，想了那么久，卧槽，这题好像有点难度，一看榜好多人切了这题，我就知道，今天想拿到名额，这题不得不切，于是我放弃所有题硬肝这题，期间队友让我想H，我把我的想法给他们后继续肝C，5个小时想了好几个线段树维护dp的假算法，导致最终大崩盘，只有两个题。赛后我用我提供给他们的想法一发切H，I题树套树更是一眼看出思路，15分钟就手写完树套树（然而tle了），E题感觉有点难度，现场我想出来的几率不大，看了中南网友的题解博客都看了好久才懂，最场比赛，就算我开局放弃C，恐怕也只能稳4题（果然如我预料，想拿名额C必切）
C. Hello 2019
cf原题，没啥好说的，链接：Good Bye 2016 E. New Year and Old Subsequence
#include<bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 10, inf = 1e9;
void up(int &x, int y) {
x = min(x, y);
}
struct mat {
int d[5][5];
mat operator+(const mat& t) const {
mat tmp;
for (int i = 0; i < 5; i++)
for (int j = 0; j < 5; j++) {
tmp.d[i][j] = inf;
for (int k = 0; k < 5; k++)
up(tmp.d[i][j], d[i][k] + t.d[k][j]);
}
return tmp;
}
} cat[maxn * 4];
#define ls o * 2
#define rs o * 2 + 1
#define mid (l + r) / 2
char s[maxn];
void build(int o, int l, int r) {
if (l == r) {
for (int i = 0; i < 5; i++)
for (int j = 0; j < 5; j++)
cat[o].d[i][j] = ((i == j) ? 0 : inf);
if (s[l] == '2')
cat[o].d[0][0] = 1, cat[o].d[0][1] = 0;
else if (s[l] == '0')
cat[o].d[1][1] = 1, cat[o].d[1][2] = 0;
else if (s[l] == '1')
cat[o].d[2][2] = 1, cat[o].d[2][3] = 0;
else if (s[l] == '9')
cat[o].d[3][3] = 1, cat[o].d[3][4] = 0;
else if (s[l] == '8')
cat[o].d[3][3] = 1, cat[o].d[4][4] = 1;
return;
}
build(ls, l, mid);
build(rs, mid + 1, r);
cat[o] = cat[ls] + cat[rs];
}
mat qu(int o, int l, int r, int ql, int qr) {
if (l >= ql && r <= qr)
return cat[o];
if (qr <= mid)
return qu(ls, l, mid, ql, qr);
else if (ql > mid)
return qu(rs, mid + 1, r, ql, qr);
else return qu(ls, l, mid, ql, qr) + qu(rs, mid + 1, r, ql, qr);
}
int main() {
int n, q, l, r;
scanf("%d%d%s", &n, &q, s + 1);
reverse(s + 1, s + 1 + n);
build(1, 1, n);
while (q--) {
scanf("%d%d", &l, &r);
l = n - l + 1;
r = n - r + 1;
swap(l, r);
int ccsu_cat = qu(1, 1, n, l, r).d[0][4];
if (ccsu_cat == inf)
ccsu_cat = -1;
printf("%d\n", ccsu_cat);
}
}
E. Magic Master
学自中南网友：019南昌网络赛 E. Magic Master
#include<bits/stdc++.h>
using namespace std;
int m, n, vis[12];
int gao(int p) {
int cat = 1, N = n - 1;
while (p % m && N > m) {
cat += N / m;
if (p / m < N / m) {
p -= p / m;
p += N % m;
}
else
p -= N / m * m;
N -= N / m;
}
if (p % m == 0)
return cat + p / m;
memset(vis, 0, sizeof(vis));
int x = 1, cnt = 0;
while (1) {
if (!vis[x]) {
if (++cnt == m) {
cat++;
if (x == p)
return cat;
vis[x] = 1;
cnt = 0;
}
}
x = x % N + 1;
}
}
int main() {
int T, q, x;
scanf("%d", &T);
while (T--) {
scanf("%d%d%d", &n, &m, &q);
m++;
while (q--) {
scanf("%d", &x);
printf("%d\n", gao(x - 1));
}
}
}
解法：我们把构建的矩阵的1到1e6的次方打表一下，1e6到1e12次方每隔1e6打表，1e12到2e18每隔1e12再打表，然后我们每次求矩阵快速幂复杂度至多2 * 矩阵相乘，举个例子，x^987654321我们拆开成 x^987000000 * x^654321即可
#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn = 1e6 + 5, mod = 998244353, T = 1e6;
struct mat {
int d[2][2];
mat operator *(const mat& t) const {
mat tmp;
for (int i = 0; i < 2; i++)
for (int j = 0; j < 2; j++)
tmp.d[i][j] = (1ll * d[i][0] * t.d[0][j] + 1ll * d[i][1] * t.d[1][j]) % mod;
return tmp;
}
};
mat a[maxn], b[maxn], c[maxn * 2], cat;
void init() {
int N = 1e6;
cat.d[0][0] = 3, cat.d[0][1] = 2, cat.d[1][0] = 1, cat.d[1][1] = 0;
a[1] = cat;
for (int i = 2; i <= N; i++)
a[i] = a[i - 1] * a[1];
b[1] = a[N];
for (int i = 2; i <= N; i++)
b[i] = b[i - 1] * b[1];
c[1] = b[N];
for (int i = 2; i <= N * 2; i++)
c[i] = c[i - 1] * c[1];
}
ll gao(ll N) {
if (N == 0)
return 0;
if (N == 1)
return 1;
N--;
mat ans;
int ccsu = 0, ccsu_cat = 0;
while (N && ccsu < 3) {
++ccsu;
int k = N % T;
if (ccsu == 3)
k = N;
if (k) {
if (ccsu == 1) {
if (ccsu_cat)
ans = ans * a[k];
else
ans = a[k];
}
else if (ccsu == 2) {
if (ccsu_cat)
ans = ans * b[k];
else
ans = b[k];
}
else {
if (ccsu_cat)
ans = ans * c[k];
else
ans = c[k];
}
ccsu_cat = 1;
}
N /= T;
}
return ans.d[0][0];
}
int main() {
init();
int q;
ll ans = 0, N;
scanf("%d%lld", &q, &N);
for (int i = 1; i <= q; i++) {
int tmp = gao(N);
ans ^= tmp;
N = N ^ (1ll * tmp * tmp);
}
printf("%lld\n", ans);
}
解法：这个题如果不带修改，用前缀和思想直接可持久化线段树即可，带修改的话，我们可以用树状数组套动态开点线段树去实现动态的前缀和，好像没啥好讲的，因为这题思维难度为0…
#include<bits/stdc++.h>
using namespace std;
#define mid (l + r) / 2
#define low(x) x&-x
const int maxn = 2e5 + 2;
int rt[maxn], ls[maxn * 210], rs[maxn * 210], sum[maxn * 210], cnt;
int a[maxn], n, V[21], tot;
void up(int l, int r, int k, int v) {
for (int i = 1; i <= tot; i++)
sum[V[i]] += v;
if (l == r)
return;
if (k <= mid) {
for (int i = 1; i <= tot; i++) {
if (!ls[V[i]])
ls[V[i]] = ++cnt;
V[i] = ls[V[i]];
}
up(l, mid, k, v);
}
else {
for (int i = 1; i <= tot; i++) {
if (!rs[V[i]])
rs[V[i]] = ++cnt;
V[i] = rs[V[i]];
}
up(mid + 1, r, k, v);
}
}
int qu(int l, int r, int ql, int qr) {
if (l >= ql && r <= qr) {
int ccsu_cat = 0;
for (int i = 1; i <= tot; i++)
ccsu_cat += sum[V[i]];
return ccsu_cat;
}
int ccsu_cat = 0;
int V2[21];
if (ql <= mid) {
for (int i = 1; i <= tot; i++)
V2[i] = V[i], V[i] = ls[V2[i]];
ccsu_cat += qu(l, mid, ql, qr);
for (int i = 1; i <= tot; i++)
V[i] = V2[i];
}
if (qr > mid) {
for (int i = 1; i <= tot; i++)
V[i] = rs[V[i]];
ccsu_cat += qu(mid + 1, r, ql, qr);
}
return ccsu_cat;
}
void modify(int x, int k, int v) {
tot = 0;
for (; x <= n; x += low(x))
V[++tot] = rt[x];
up(1, n, k, v);
}
int query(int x, int ql, int qr) {
tot = 0;
for (; x; x -= low(x))
V[++tot] = rt[x];
return qu(1, n, ql, qr);
}
int main() {
int q, opt, l, r, x, y;
scanf("%d%d", &n, &q);
for (int i = 1; i <= n; i++)
rt[i] = ++cnt;
for (int i = 1; i <= n; i++) {
scanf("%d", &a[i]);
if (a[i] != a[i - 1])
modify(i, a[i], 1);
}
while (q--) {
scanf("%d%d%d", &opt, &l, &r);
if (opt == 1) {
if (a[l] == r)
continue;
if (a[l] != a[l - 1])
modify(l, a[l], -1);
if (a[l] == a[l + 1])
modify(l + 1, a[l], 1);
else if (r == a[l + 1])
modify(l + 1, a[l + 1], -1);
a[l] = r;
if (a[l] != a[l - 1])
modify(l, a[l], 1);
}
else {
scanf("%d%d", &x, &y);
int cat = query(r, x, y) - query(l - 1, x, y);
if (a[l] == a[l - 1] && a[l] >= x && a[l] <= y)
cat++;
printf("%d\n", cat);
}
}
}
cdq 分治代码：
#include <bits/stdc++.h>
using namespace std;
#define mid (l + r) / 2
#define low(x) x & -x
const int maxn = 2e5 + 10;
struct node
{
int opt, x, y, v, id;
bool operator<(const node &t) const
{
return x <= t.x;
}
} cat[maxn * 5], jie[maxn * 5];
int a[maxn], ans[maxn], c[maxn], n;
void up(int x, int v)
{
for (; x <= n; x += low(x))
if (!v)
c[x] = v;
else
c[x] += v;
}
int qu(int x)
{
int res = 0;
for (; x; x -= low(x))
res += c[x];
return res;
}
void cdq(int l, int r)
{
if (l == r)
return;
cdq(l, mid);
cdq(mid + 1, r);
int p = l, q = mid + 1, T = l;
while (p <= mid && q <= r)
{
if (jie[p] < jie[q])
{
if (jie[p].opt == 0)
up(jie[p].y, jie[p].v);
cat[T++] = jie[p++];
}
else
{
if (jie[q].opt != 0)
ans[jie[q].id] += jie[q].opt * (qu(jie[q].v) - qu(jie[q].y - 1));
cat[T++] = jie[q++];
}
}
while (p <= mid)
cat[T++] = jie[p++];
while (q <= r)
{
if (jie[q].opt != 0)
ans[jie[q].id] += jie[q].opt * (qu(jie[q].v) - qu(jie[q].y - 1));
cat[T++] = jie[q++];
}
for (int i = l; i <= mid; i++)
if (jie[i].opt == 0)
up(jie[i].y, 0);
for (int i = l; i <= r; i++)
jie[i] = cat[i];
}
int main()
{
int q, opt, l, r, x, y, pos, v, tot = 0;
scanf("%d%d", &n, &q);
for (int i = 1; i <= n; i++)
{
scanf("%d", &a[i]);
if (a[i] != a[i - 1])
jie[++tot] = {0, i, a[i], 1, 0};
}
int cnt = 0;
for (int i = 1; i <= q; i++)
{
scanf("%d", &opt);
if (opt == 1)
{
scanf("%d%d", &pos, &v);
if (a[pos] == v)
continue;
if (pos > 1 && a[pos] != a[pos - 1])
jie[++tot] = {0, pos, a[pos], -1, 0};
if (pos < n && a[pos] != a[pos + 1])
jie[++tot] = {0, pos + 1, a[pos + 1], -1, 0};
a[pos] = v;
if (pos > 1 && a[pos] != a[pos - 1])
jie[++tot] = {0, pos, a[pos], 1, 0};
if (pos < n && a[pos] != a[pos + 1])
jie[++tot] = {0, pos + 1, a[pos + 1], 1, 0};
}
else
{
scanf("%d%d%d%d", &l, &r, &x, &y);
jie[++tot] = {1, r, x, y, ++cnt};
jie[++tot] = {-1, l, x, y, cnt};
if (a[l] >= x && a[l] <= y)
ans[cnt] = 1;
}
}
cdq(1, tot);
for (int i = 1; i <= cnt; i++)
printf("%d\n", ans[i]);
}
