写在前面
对于这个机制提出的建议：
-
题目偏多，偏杂。建议取消硬性规定题目，这种大规模刷题应该放在平时，而不是放在NOIP这种大考前，考前主要是找自己薄弱点
-
每周写博客的机制可以保留，不一定大家都要写一样的一套题目，毕竟每个人都有自己的薄弱处，可以刻意刷一些自己不熟练的题目，然后每周写成博客，方便参考
题解
A(模拟)
大模拟，儒略日100一轮回来取模，1852年特殊处理一下，1853年之后400年一轮回取模
之前写了一个400 + 100 + 4 的奇怪轮回，但是后面100与4的轮回完全没必要，反而增加了代码难度
tips：在能过的前提下，代码越简单越好
#include <bits/stdc++.h>
#define int long long
using namespace std;
int JLtotDays = 0;
int T, Year, Month, Day, Res1582;
int Year400tot, Year100tot, Year18tot; // common
int month[13] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int read () {
int tot = 0, f = 1; char c = getchar ();
while (c < '0' || c > '9') { if (c == '-') f = -1; c = getchar (); }
while (c >= '0' && c <= '9') { tot = tot * 10 + c - '0'; c = getchar (); }
return tot * f;
}
signed main () {
// freopen ("P7075_6.in", "r", stdin);
// freopen ("julian.out", "w", stdout);
for (int i = -4712; i < 1582; i++) {
JLtotDays += 365;
if (i % 4 == 0) JLtotDays++;
}
for (int i = 1; i < 10; i++) JLtotDays += month[i]; JLtotDays += 4;
Res1582 = 31-15 + month[11] + month[12];
for (int i = 1583; i < 1983; i++) {
Year400tot += 365;
if (i % 4 == 0 && i % 100 != 0) Year400tot++;
else if (i % 400 == 0) Year400tot++;
}
for (int i = 1583; i < 1683; i++) {
Year100tot += 365;
if (i % 4 == 0 && i % 100 != 0) Year100tot++;
} // 预处理
for (int i = 1583; i <= 1600; i++) {
Year18tot += 365;
if (i % 4 == 0 && i % 100 != 0) Year18tot++;
}
// cout<<JLtotDays<<" "<<Res1582<<endl;
T = read ();
while (T--) {
month[2] = 28;
Day = read (); Day++;
if (Day == 2299162) { // 特判
printf ("15 10 1582
");
continue;
}
if (Day <= JLtotDays) { // 儒略日
Year = -4712;
if (Day > 4 * 365 + 1) {
Year += 4 * (Day / (4 * 365 + 1));
Day %= (4 * 365 + 1);
if (Day == 0) Day += (4 * 365 + 1), Year -= 4;
}
// cout<<Year<<endl;
bool flag = 0;
if (Day > 366) Year++, Day -= 366, flag = 1;
if (Day > 365 && flag) Year += (Day / 365), Day %= 365;
if (Day == 0) Day += 365, Year--;
if (Year % 4 == 0) month[2] = 29;
for (Month = 1; Day > month[Month]; Month++) Day -= month[Month];
printf ("%d %d ", Day, Month);
if (Year <= 0) {
printf ("%d BC
", -Year + 1);
}
else printf ("%d
", Year);
}
else { Day -= JLtotDays; // 格里高利历
if (Day <= Res1582) {
if (Day <= 31 - 15)
printf ("%d 10 1582
", Day + 15);
else if (Day <= 31 - 15 + month[11])
printf ("%d 11 1582
", Day - (31 - 15));
else printf ("%d 12 1582
", Day -(31 - 15 + month[11]));
}
else {
Day -= Res1582;
Year = 1583;
if (Day > Year400tot) {
Year += 400 * (Day / Year400tot);
Day %= Year400tot;
if (Day == 0) Year -= 400, Day += Year400tot;
}
bool flag = 0, tag = 0;
if (Day > Year100tot + 1)
Day -= Year100tot + 1, Year += 100, flag = 1;
if (Day > Year100tot && flag) {
Year += 100 * (Day / Year100tot), Day %= Year100tot;
if (Day == 0) Year -= 100, Day += Year100tot;
}
// cout<<Day<<" "<<Year<<" "<<Year18tot<<endl;
if (Day > Year18tot) {
if ((Year + 17) % 400 == 0 ) {
if (Day > Year18tot + 1) Day -= Year18tot + 1, Year += 18, tag = 1;
}
else Day -= Year18tot, Year += 18; tag = 1;
}
// cout<<Day<<" "<<Year<<" "<<tag<<endl;
if (Day > (4 * 365 + 1)) {
Year += 4 * (Day / (4 * 365 + 1));
Day %= (4 * 365 + 1);
if (Day == 0) Year -= 4, Day += 4 * 365 + 1;
}
// cout<<Day<<" "<<Year<<" "<<tag<<endl;
if (!tag) {
flag = 0;
if (Day > 365) Year ++, Day -= 365;
if (Day > 366) Year ++, Day -= 366, flag = 1;
if (Day > 365 && flag) Year ++, Day -= 365;
}
else if (Day > 365) {
int t = 365; if ((Year % 4 == 0 && Year % 100 != 0) || Year % 400 == 0) t++;
while (Day > t) {
Day -= t; Year ++;
t = 365;
if ((Year % 4 == 0 && Year % 100 != 0) || Year % 400 == 0) t++;
}
}
// cout<<Day<<endl;
if ((Year % 4 == 0 && Year % 100 != 0) || Year % 400 == 0) month[2] = 29;
for (Month = 1; Day > month[Month] && Month < 12; Month++) Day -= month[Month];
/*if (Day == 0 && Month == 1) {
Month = 12, Year--, Day = 31;
if ((Year % 4 == 0 && Year % 100 != 0) || Year % 400 == 0) Day = 30;
}*/
// if (Day == 1 && Month == 1 && (((Year - 1) % 4 == 0 && (Year - 1) % 100 != 0) || (Year - 1) % 400 == 0)) Day = 31, Month = 12, Year--;
printf ("%d %d %d
", Day, Month, Year);
}
}
}
return 0;
}
B(位运算)
把所有给出的动物全部或起来，这样就得到了一个包含所有初始动物所需的饲料的序列，之后直接算一下就得出答案了
注意：会爆ull，可以分两次加，先一次，再减去个n，之后再加
#include <bits/stdc++.h>
#define int unsigned long long
using namespace std;
const int K = 65;
int n, m, k, c, a, q, p;
bool v[K];
int s;
inline int read () {
int tot = 0, f = 1; char c = getchar ();
while (c < '0' || c > '9') { if (c == '-') f = -1; c = getchar (); }
while (c >= '0' && c <= '9') { tot = tot * 10 + c - '0'; c = getchar (); }
return tot * f;
}
signed main () {
// freopen ("zoo3.in", "r", stdin);
n = read (); m = read (); c = read (); k = read ();
for (int i = 1; i <= n; i++) {
a = read (); s |= a;
}
for (int i = 1; i <= m; i++) {
p = read (); q = read ();
v[p]++;
}
int ans = 0;
for (int i = 0; i < k; i++) {
if (s >> i&1 || !v[i]) ans++;
}
if (ans == 0) {
printf ("0
");
return 0;
}
if (ans == 64 && n == 0) {
cout<<"18446744073709551616
";
return 0;
}
int Ans = (1ull << (ans - 1));
cout << Ans - n + Ans << endl; // 就像这样
return 0;
}
C(拓扑排序，建图)
题目给出三个操作，单点加，全部乘，和调用几个函数。
我们把所有的操作都看成一个个节点，根据调用关系建出一张图，因为保证不存在有递归，所以调用关系一定是一个DAG。
做出贡献的点一定是出度为0的点，其他的点只是反复调用它们
当我们使用乘这个操作时，会把前面的所有操作都乘上一个数，我们就可以逆着拓扑求一遍每个节点产生“乘”这个操作的贡献是多少
同理，我们可以处理加法，由于处理乘法是为了得到后面的状态才去逆着推，但是加法需要得到当前的值，所以我们需要顺推
此时，我们把乘法标记下放，然后先乘再加，即可得到当前的值
#include <bits/stdc++.h>
#define int long long
using namespace std;
const int N = 3e6, MOD = 998244353;
struct Node {
int next, to;
} edge[N];
int head[N], in[N], cnt;
int ans[N], P[N], V[N], I[N], T[N], C[N];
int n, m, q;
int Q[N], r;
inline int read () {
int tot = 0, f = 1; char c = getchar ();
while (c < '0' || c > '9') { if (c == '-') f = -1; c = getchar (); }
while (c >= '0' && c <= '9') { tot = tot * 10 + c - '0'; c = getchar (); }
return tot * f;
}
inline void add (int x, int y) {
edge[++cnt].next = head[x];
edge[cnt].to = y;
head[x] = cnt;
in[y]++;
}
inline int calc (int u) {
if (T[u] != -1) return T[u]; T[u] = 1;
for (int i = head[u]; i; i = edge[i].next)
T[u] = T[u] * calc (edge[i].to) % MOD;
return T[u];
}
signed main () {
n = read ();
for (int i = 1; i <= n; i++) add (0, i), P[i] = i, V[i] = read (), I[i] = 1, T[i] = 1;
m = read ();
for (int i = n + 1; i <= n + m; i++) {
I[i] = read ();
if (I[i] == 1) P[i] = read (), V[i] = read (), T[i] = 1;
else if (I[i] == 2) T[i] = V[i] = read ();
else { int k = read ();
for (int j = 1; j <= k; j++) add (i, read() + n), T[i] = -1;
}
}
for (int i = 1; i <= n + m; i++) calc (i);
q = read ();
for (int i = 1; i <= q; i++) add (0, read () + n);
for (int i = 0; i <= n + m; i++) if (!in[i]) Q[++r] = i;
I[0] = 3, C[0] = 1;
while (r) {
int u = Q[r], x = C[u]; r--;
if (I[u] == 3) {
for (int i = head[u]; i; i = edge[i].next) {
int v = edge[i].to;
C[v] = (x + C[v]) % MOD; in[v]--;
if (in[v] == 0) Q[++r] = v; x = x * T[v] % MOD;
}
}
else if (I[u] == 1) ans[P[u]] = (1 * V[u] * x + ans[P[u]]) % MOD;
}
for (int i = 1; i <= n; i++) printf ("%d ", ans[i]);
return 0;
}
感觉这个好看点
#include<cstdio>
#include<cstring>
#include<vector>
#include<algorithm>
using namespace std;
const int N=1e5+9,mod=998244353;
vector<int> G[N],g[N];
int n,m,q,mul[N],op[N],add[N],f[N];
int Q[N],hh,tt,a[N],pos[N],in[N];
int Add[N],Mul[N];
inline int read()
{
int res=0,f=1;
char ch=getchar();
while(ch<'0'||ch>'9')
{
if(ch=='-') f=-1;
ch=getchar();
}
while(ch>='0'&&ch<='9')
res=(res<<3)+(res<<1)+ch-'0',ch=getchar();
return res*f;
}
inline void rev_toposort()
{
hh=0,tt=-1;
for(int i=1;i<=m;i++) if(!in[i]) Q[++tt]=i;
while(hh<=tt)
{
int u=Q[hh++];
for(int i=0;i<g[u].size();i++)
{
int v=g[u][i];
mul[v]=1ll*mul[u]*mul[v]%mod;
if(--in[v]==0) Q[++tt]=v;
}
}
}
inline void toposort()
{
memset(in,0,sizeof in);
for(int i=1;i<=m;i++)
for(int j=0;j<G[i].size();j++)
in[G[i][j]]++;
hh=0,tt=-1;
for(int i=1;i<=m;i++) if(!in[i]) Q[++tt]=i;
while(hh<=tt)
{
int u=Q[hh++];
for(int i=G[u].size()-1;i>=0;i--)
{
int v=G[u][i];
Add[v]=(Add[v]+Add[u])%mod;
Add[u]=1ll*Add[u]*mul[v]%mod;
if(--in[v]==0) Q[++tt]=v;
}
}
}
int main()
{
// freopen("call.in","r",stdin);
// freopen("call.out","w",stdout);
n=read();
for(int i=1;i<=n;i++) a[i]=read();
m=read();
for(int i=1;i<N;i++) mul[i]=1;
for(int i=1;i<=m;i++)
{
op[i]=read();
if(op[i]==1) pos[i]=read(),add[i]=read();
if(op[i]==2) mul[i]=read();
if(op[i]==3)
{
int cnt;cnt=read();
while(cnt--)
{
int x;x=read();
g[x].push_back(i);in[i]++;
G[i].push_back(x);
}
}
}
rev_toposort();
q=read();
for(int i=1;i<=q;i++) f[i]=read();Mul[q+1]=1;
for(int i=q;i>=1;i--)
{
Mul[i]=1ll*Mul[i+1]*mul[f[i]]%mod;
Add[f[i]]=(Add[f[i]]+Mul[i+1])%mod;
}
for(int i=1;i<=n;i++) a[i]=1ll*a[i]*Mul[1]%mod;
toposort();
for(int i=1;i<=m;i++)
if(op[i]==1) a[pos[i]]=(a[pos[i]]+1ll*add[i]*Add[i]%mod)%mod;//printf("%d %d
",pos[i],Add[i]);
for(int i=1;i<=n;i++)
printf("%d ",a[i]);
return 0;
}
D
我们来转换一下提意，题目就是给出一些蛇，从大到小进行决策“该不该吃最小的蛇”，并且每一条蛇都必须保证自己不会被吃掉。
首先，我们可以得出一个结论：凡是做出过选择的蛇都会活到最后。
那么我们就先假装所有的蛇都很蠢，只要自己不会最短就吃，但是即使成为了最短的蛇，也不一定会被吃，因为当时最长的蛇为了顾全大局会不敢冒险吃你，所以就存在一种奇偶性，因为后蛇不敢冒险，所以前蛇就敢吃，然后之后蛇又不敢冒险。
用两个双端队列，一个存做出决策的蛇，一个存还没有做出决策的蛇
#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 5;
int T, n;
deque <int> q1, q2;
int a[N], val[N];
inline int read () {
int tot = 0, f = 1; char c = getchar ();
while (c < '0' || c > '9') { if (c == '-') f = -1; c = getchar (); }
while (c >= '0' && c <= '9') { tot = tot * 10 + c - '0'; c = getchar (); }
return tot * f;
}
inline void init () {
while (q1.size ()) q1.pop_back ();
while (q2.size ()) q2.pop_back ();
for (int i = 1; i <= n; i++) val[i] = a[i];
q1.push_front(n);
for (int i = 1; i < n; i++) q2.push_back (i);
}
inline int can (int x, int y) {
if (x == y) return 0;
if (val[x] < val[y]) return -1;
if (val[x] > val[y]) return 1;
if (x < y) return -1;
if (x > y) return 1;
return 0;
}
inline bool check () {
if (q1.size () == 1) return 0;
if (q1.size () == 2) return 1;
int x = q1.front(); q1.pop_front();
int y = q1.back(); q1.pop_back();
val[y] -= val[x];
if (can(y, q1.front()) < 0) {
q1.push_front (y);
return !check();
}
return 1;
}
inline bool calc () {
if (q2.size () == 0) return 0;
int siz = q1.size () + q2.size ();
int x = q1.back(); q1.pop_back ();
int y = q2.front(); q2.pop_front();
val[x] -= val[y];
while (q2.size () && can (x, q2.back ()) < 0) {
q1.push_front (q2.back ());
q2.pop_back ();
} q1.push_front (x);
if (q2.size ()) return 1;
if (!check()) return 1;
return 0;
}
inline int solve () {
init ();
int ans = n;
while (calc ()) {
ans --;
// cout<<ans<<endl;
}
return ans;
}
signed main () {
T = read () - 1; n = read ();
for (int i = 1; i <= n; i++) a[i] = read ();
printf ("%d
", solve());
while (T--) {
int k = read ();
for (int i = 1; i <= k; i++) {
int x = read (), y = read ();
a[x] = y;
}
printf ("%d
", solve());
}
return 0;
}
E
F
G
H
I
J(莫队，栈)
大概意思就是说有一些数字，有一些询问，每次询问某个区间内只出现一次的数，如果有多个给出一个即可，支持离线
直接用莫队，开个栈记录一下只出现一次的元素
#include <bits/stdc++.h>
using namespace std;
const int N = 5e5 + 5;
struct Node {
int l, r, id;
} Ask[N];
int block, a[N], belong[N];
int n, q, sta[N], pos[N], ans[N], cnt[N];
int top, l, r;
inline int read () {
int tot = 0, f = 1; char c = getchar ();
while (c < '0' || c > '9') { if (c == '-') f = -1; c = getchar (); }
while (c >= '0' && c <= '9') { tot = tot * 10 + c - '0'; c = getchar (); }
return tot * f;
}
inline bool cmp (Node x, Node y) {
if (belong[x.l] ^ belong[y.l]) return x.l < y.l;
else if (belong[x.l] & 1) return x.r < y.r;
else return x.r > y.r;
}
inline void add (int t) {
cnt[t]++;
if (cnt[t] == 1) {
sta[++top] = t;
pos[t] = top;
}
else if (cnt[t] == 2) {
sta[pos[t]] = sta[top];
pos[sta[top]] = pos[t];
sta[top--] = pos[t] = 0;
}
}
inline void del (int t) {
cnt[t]--;
if (cnt[t] == 1) {
sta[++top] = t;
pos[t] = top;
}
else if (cnt[t] == 0) {
sta[pos[t]] = sta[top];
pos[sta[top]] = pos[t];
sta[top--] = pos[t] = 0;
}
}
signed main () {
n = read (); block = sqrt(n);
for (int i = 1; i <= n; i++) a[i] = read (), belong[i] = i / block + 1;
q = read ();
for (int i = 1; i <= q; i++) {
Ask[i].l = read (); Ask[i].r = read ();
Ask[i].id = i;
}
sort (Ask + 1, Ask + 1 + q, cmp);
/*for (int i = 1; i <= q; i++) {
printf ("%d %d
", Ask[i].l, Ask[i].r);
}*/
l = r = 1;
add (a[1]);
for (int i = 1; i <= q; i++) {
while (r < Ask[i].r) add (a[++r]);
while (r > Ask[i].r) del (a[r--]);
while (l < Ask[i].l) del (a[l++]);
while (l > Ask[i].l) add (a[--l]);
ans[Ask[i].id] = sta[top];
}
for (int i = 1; i <= q; i++)
printf ("%d
", ans[i]);
return 0;
}
K(模拟)
显然......
#include <bits/stdc++.h>
using namespace std;
int n, d;
int ans, lax, x;
inline int read () {
int tot = 0, f = 1; char c = getchar ();
while (c < '0' || c > '9') { if (c == '-') f = -1; c = getchar (); }
while (c >= '0' && c <= '9') { tot = tot * 10 + c - '0'; c = getchar (); }
return tot * f;
}
signed main () {
n = read (); d = read (); lax = read ();
for (int i = 2; i <= n; i++) {
x = read (); if (x - lax == d * 2) ans++;
else if (x - lax > d * 2) ans += 2;
lax = x;
}
printf ("%d
", ans + 2);
return 0;
}
L(模拟)
显然01交错最优
#include <bits/stdc++.h>
using namespace std;
int n, m;
signed main () {
cin >> n;
for (int i = 1; i <= n; i++)
cout<<i % 2;
cout<<endl;
return 0;
}