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"Team training competition" Shandong multi university training 3
2022-06-30 22:20:00 【Chels.】
A - Ant colony
Title Description ：
Here you are. n Number
a[i],m Time to ask , Give one at a timel, r, ask[l, r]How many numbers are there ina[i]dissatisfaction ：a[i]It can be divided by all numbers in the interval
Ideas ：
An analysis shows that , If a number in an interval can be divided by all other numbers , Then this number must be equal to all the numbers in the interval gcd After the results of the
So we need a data structure that can maintain intervals gcd, And maintain intervals gcd The number of
Here are two ways ：
- st surface + Two points , use
stTable maintenance intervalgcd, When querying the quantity , We can use one vector Array to store where each number appears , Just use two points to check , But because the number is 1e9, Out-of-service vector Array , So it can be usedmap<int, vector>Or write a discretization and then vector Array storage- Segment tree maintains intervals gcd Values and intervals of gcd Number , It is a simple ordinary segment tree without modification
#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
#define inf 0x3f3f3f3f
#define mod 998244353
#define m_p(a,b) make_pair(a, b)
#define mem(a,b) memset((a),(b),sizeof(a))
#define io ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
typedef long long ll;
typedef pair <int,int> pii;
#define MAX 300000 + 50
int n, m, k, op;
int l, r;
int tr[MAX];
int lg[MAX];
int gg[MAX][30];
map<int, vector<int>>mp;
int gcd(int a, int b){
if(b) return gcd(b, a % b);
else return a;
}
void init(){
for(int i = 1; i <= n; ++i)gg[i][0] = tr[i];
for(int i = 2; i <= n; ++i){
lg[i] = lg[i / 2] + 1;
}
for(int j = 1; j <= lg[n]; ++j){
for(int i = 1; i + (1 << j) - 1 <= n; ++i){
gg[i][j] = gcd(gg[i][j - 1], gg[i + (1 << (j - 1))][j - 1]);
}
}
}
int getgcd(int l, int r){
int len = lg[r - l + 1];
return gcd(gg[l][len], gg[r - (1 << len) + 1][len]);
}
void work(){
cin >> n;
for(int i = 1; i <= n; ++i){
cin >> tr[i];
mp[tr[i]].push_back(i);
}
init();
cin >> m;
while(m--){
cin >> l >> r;
int x = getgcd(l, r);
cout << r - l + 1 - (upper_bound(mp[x].begin(), mp[x].end(), r) - lower_bound(mp[x].begin(), mp[x].end(), l))<<endl;
}
}
int main(){
io;
work();
return 0;
}
#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
#define inf 0x3f3f3f3f
#define mod 998244353
#define m_p(a,b) make_pair(a, b)
#define mem(a,b) memset((a),(b),sizeof(a))
#define io ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#define ls (p<<1)
#define rs ((p<<1)|1)
typedef long long ll;
typedef pair <bool,int> pii;
#define MAX 500000 + 50
int n, m, k, op;
int tr[MAX];
int num[MAX];
int gg[MAX];
int gcd(int a, int b){
if(b) return gcd(b, a % b);
else return a;
}
struct ran{
int gg, num;
};
void pushup(int p){
gg[p] = gcd(gg[ls], gg[rs]);
num[p] = (gg[ls] == gg[p] ? num[ls] : 0) + (gg[rs] == gg[p] ? num[rs] : 0);
}
void built(int p, int l, int r){
if(l == r){
gg[p] = tr[l];
num[p] = 1;
return;
}
int mid = (l + r) / 2;
built(ls, l, mid);
built(rs, mid + 1, r);
pushup(p);
}
ran getans(int p, int l, int r, int s, int t){
if(s <= l && r <= t){
return {
gg[p], num[p]};
}
ran ans = {
0, 0};
int mid = (l + r) / 2;
if(mid >= s){
ans = getans(ls, l, mid, s, t);
}
if(mid < t){
ran a = ans;
ran now = getans(rs, mid + 1, r, s, t);
ans.gg = gcd(ans.gg, now.gg);
ans.num = (a.gg == ans.gg ? a.num : 0) + (now.gg == ans.gg ? now.num : 0);
}
return ans;
}
void work(){
cin >> n;
for(int i = 1; i <= n; ++i)cin >> tr[i];
built(1, 1, n);
cin >> m;
int l, r;
while(m--){
cin >> l >> r;
ran ans = getans(1, 1, n, l, r);
cout << r - l + 1 - ans.num << endl;
}
}
int main(){
io;
work();
return 0;
}
B - New String
Title Description ：
Give it to you again 26 The order of letters , Redefine the dictionary order in this order , Here you are.
nA string , Output NokSmall strings
Ideas ：
According to the given 26 The first letter is map Mapping the , Then write a string comparison function sort Just output it later
#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
#define inf 0x3f3f3f3f
#define mod 998244353
#define m_p(a,b) make_pair(a, b)
#define mem(a,b) memset((a),(b),sizeof(a))
#define io ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#define ls (p<<1)
#define rs ((p<<1)|1)
typedef long long ll;
typedef pair <bool,int> pii;
#define MAX 300000 + 50
int n, m, k, op;
string s;
map<char, int>mp;
struct ran{
string s;
bool operator < (const ran &x)const{
for(int i = 0; i < s.size() && i < x.s.size(); ++i){
if(mp[s[i]] < mp[x.s[i]])return true;
else if(mp[s[i]] > mp[x.s[i]])return false;
else continue;
}
return s.size() < x.s.size();
}
}tr[MAX];
void work(){
cin >> s;
for(int i = 0; i < s.size(); ++i){
mp[s[i]] = i + 1;
}
cin >> n;
for(int i = 1; i <= n; ++i){
cin >> tr[i].s;
}
sort(tr + 1, tr + 1 + n);
cin >> k;
cout << tr[k].s << endl;
}
int main(){
io;
work();
return 0;
}
C - Easy Problem
Title Description ：
Here you are.
n * mGraph ,*It's an obstacle ,.It's open spaceNow there are two people in
a,bThe location of , They move in the same direction , Will rise at the same time 、 Next 、 Left 、 Move in four right directions , If you move in a certain direction , Encountering obstacles or boundaries , Then this person will stay where he is , But another person can walk normally , Of course, if you encounter obstacles in the same direction, you can't move towards that thing , Ask when they can meet
Ideas ：
If you look at it, you can see that
nonly 50, It's only a matter of running all over the map 504 The situation of , Very small , Run straightbfsJust go
#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
#define inf 0x3f3f3f3f
#define mod7 1000000007
#define mod9 998244353
#define m_p(a,b) make_pair(a, b)
#define mem(a,b) memset((a),(b),sizeof(a))
#define io ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#define debug(a) cout << "Debuging...|" << #a << ": " << a << "\n";
typedef long long ll;
typedef pair <int,int> pii;
#define y1 y114514
#define MAX 300000 + 50
int n, m, k, x;
char tr[55][55];
bool vis[55][55][55][55];
int dx[] = {
1, 0, -1, 0};
int dy[] = {
0, 1, 0, -1};
int x1, y1, x2, y2;
struct ran{
int x1, y1, x2, y2, d;
};
bool judge(int x, int y){
if(x < 1 || x > n || y < 1 || y > n)return false;
if(tr[x][y] == '*')return false;
else return true;
}
void bfs(){
queue<ran>q;
ran now, nex;
q.push({
x1, y1, x2, y2, 0});
vis[x1][y1][x2][y2] = 1;
while(!q.empty()){
now = q.front();q.pop();
if(now.x1 == now.x2 && now.y114514 == now.y2){
cout << now.d << endl;
return;
}
for(int i = 0; i < 4; ++i){
if(judge(now.x1 + dx[i], now.y114514 + dy[i])){
nex.x1 = now.x1 + dx[i];
nex.y114514 = now.y114514 + dy[i];
}
else {
nex.x1 = now.x1;
nex.y114514 = now.y114514;
}
if(judge(now.x2 + dx[i], now.y2 + dy[i])){
nex.x2 = now.x2 + dx[i];
nex.y2 = now.y2 + dy[i];
}
else{
nex.x2 = now.x2;
nex.y2 = now.y2;
}
if(vis[nex.x1][nex.y114514][nex.x2][nex.y2])continue;
nex.d = now.d + 1;
q.push(nex);
vis[nex.x1][nex.y114514][nex.x2][nex.y2] = 1;
}
}
cout << "no solution\n";
}
void work(){
cin >> n;
for(int i = 1; i <= n; ++i){
for(int j = 1; j <= n; ++j){
cin >> tr[i][j];
if(tr[i][j] == 'a'){
tr[i][j] = '.';
x1 = i;y1 = j;
}
else if(tr[i][j] == 'b'){
tr[i][j] = '.';
x2 = i;y2 = j;
}
}
}
bfs();
}
int main(){
io;
work();
return 0;
}
D - Maximum Value
Title Description ：
Here you are. n Number , choose
i, j, And meeta[j] >= a[i], aska[j] % a[i]The maximum of
Ideas ：
We can sort the array first , After sorting , For a number
a[i], We need to find onea[j], seeka[j]%a[i]The maximum of ,a[j] = k * a[i] + x,xThat is, the remainder , We can enumerate k, Then find the first greater than or equal to in the arrayk*a[i]The location ofp, bep-1The position should be(k-1)*a[i] To k * a[i]thisa[i]The one with the largest remainder , For all thek, We all calculate the remainder to find a maximum , This is the correspondinga[i]The remainder that can be produced as a divisor , We treat eachiIt's OK to ask for it once , The complexity should be harmonic series ,O(nlogn)Pay attention to pruning , That is, just look for the same number once
#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
#define inf 0x3f3f3f3f
#define mod 998244353
#define m_p(a,b) make_pair(a, b)
#define mem(a,b) memset((a),(b),sizeof(a))
#define io ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
typedef long long ll;
typedef pair <int,int> pii;
#define MAX 300000 + 50
int n, m, k, op;
int tr[MAX];
void work(){
cin >> n;
for(int i = 1; i <= n; ++i){
cin >> tr[i];
}
sort(tr + 1, tr + 1 + n);
int ans = 0;
for(int i = 1; i <= n; ++i){
int x = tr[i];
if(tr[i] == tr[i - 1])continue;
if(x == 1)continue;
while(x - tr[i] <= tr[n]){
int p = (int)(lower_bound(tr + i + 1, tr + 1 + n, x) - tr);
ans = max(ans, tr[p - 1] % tr[i]);
x += tr[i];
}
}
cout << ans << endl;
}
int main(){
io;
work();
return 0;
}
E - Prefix Equality
Title Description ：
Here are two arrays
ar[i], br[i],Q Time to ask , Each inquiry givesa, b, askar[1] To ar[a]A set of numbers andbr[1] To br[b]Whether the set composed of numbers of is the same
Ideas ：
Just hash it , I made three kinds of arrays , An array of quantities , An array of prefixes and , An array of prefix products , When judging, the three are the same
#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
#define inf 0x3f3f3f3f
#define mod7 1000000007
#define mod9 998244353
#define m_p(a,b) make_pair(a, b)
#define mem(a,b) memset((a),(b),sizeof(a))
#define io ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#define debug(a) cout << "Debuging...|" << #a << ": " << a << "\n";
typedef long long ll;
typedef pair <int,int> pii;
#define MAX 300000 + 50
int n, m, k, x, y;
ll ar[MAX], br[MAX];
ll sum1[MAX], sum2[MAX];
ll mul1[MAX], mul2[MAX];
ll num1[MAX], num2[MAX];
void work(){
cin >> n;
mul1[0] = mul2[0] = 1;
for(int i = 1; i <= n; ++i)cin >> ar[i];
for(int i = 1; i <= n; ++i)cin >> br[i];
set<ll>se;
for(int i = 1; i <= n; ++i){
sum1[i] = sum1[i - 1];
mul1[i] = mul1[i - 1];
num1[i] = num1[i - 1];
if(!se.count(ar[i])){
++num1[i];
se.insert(ar[i]);
sum1[i] += ar[i];
(mul1[i] *= ar[i]) %= mod7;
}
}
se.clear();
for(int i = 1; i <= n; ++i){
sum2[i] = sum2[i - 1];
mul2[i] = mul2[i - 1];
num2[i] = num2[i - 1];
if(!se.count(br[i])){
++num2[i];
se.insert(br[i]);
sum2[i] += br[i];
(mul2[i] *= br[i]) %= mod7;
}
}
cin >> m;
while (m--) {
cin >> x >> y;
if(num1[x] == num2[y] && sum1[x] == sum2[y] &&
mul1[x] == mul2[y])cout << "Yes\n";
else cout << "No\n";
}
}
int main(){
io;
work();
return 0;
}
G - Optimal Biking Strategy
Title Description ：
You are now 0 The location of , Need to get to
pThe location of riceYes
nA bicycle station , You can only get on and off at the bicycle stop , You can ride at most for every dollar you spendsrice , You can't ride more than one meter , If the current vehicle cannot support riding from this site to the next site , Then he can't rideAsk how many meters you need to walk on the ground at least
Ideas ：
Dynamic programming
dp[i][j]ToiThe position of meters , It tookjThe longest distance yuan can rideState transition is also good to think about , Just enumerate the last flowers
kYuan arrivedjLet's assume it took
kelement , Then we need to know where to spendkYuan Neng walked as far as possible to reachj, Make a note of this positionid, bedp[i][j] = max(dp[i][j - k] + tr[j] - tr[id];And ask
idThe method is dichotomyBe careful not to forget to add
dp[i][j] = min(dp[i][j], dp[i - 1][j])
#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
#define inf 0x3f3f3f3f
#define mod7 1000000007
#define mod9 998244353
#define m_p(a,b) make_pair(a, b)
#define mem(a,b) memset((a),(b),sizeof(a))
#define io ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#define debug(a) cout << "Debuging...|" << #a << ": " << a << "\n";
typedef long long ll;
typedef pair <int,int> pii;
#define MAX 300000 + 50
ll n, kk;
ll p, s;
ll tr[MAX];
ll id[MAX][7];
ll dp[MAX][7];
void work(){
scanf("%lld%lld%lld", &n, &p, &s);
for(int i = 1; i <= n; ++i){
scanf("%lld", &tr[i]);
}
scanf("%lld", &kk);
for(int i = 1; i <= n; ++i){
id[i][0] = i;
for(int j = 1; j <= kk; ++j){
ll pos = tr[i] - j * s;
id[i][j] = lower_bound(tr + 1, tr + 1 + n, pos) - tr;
}
}
ll ans = 1e18;
for(int i = 1; i <= n; ++i){
dp[i][0] = 0;
for(int j = 1; j <= kk; ++j){
dp[i][j] = dp[i - 1][j];
for(int k = 0; k <= j; ++k){
dp[i][j] = max(dp[i][j], dp[id[i][k]][j - k] + tr[i] - tr[id[i][k]]);
}
}
}
for(int i = 1; i <= n; ++i){
ans = min(ans, p - dp[i][kk]);
}
cout << ans << endl;
}
int main(){
io;
work();
return 0;
}
I - 250-like Number
Title Description ：
k = p * q * q * q,p,qAll prime numbers , Ask less than or equal tonHow many of them k
Ideas ：
First use the Euler sieve to screen out 1e6 The prime number in , Enumerate each
p, Calculate by two pointsqThe number ofWhat needs attention is judgment
p*q*q*q <= nWill explode whenlong long, So we can move items , Judgmentq*q*q <= n / p
#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
#define inf 0x3f3f3f3f
#define mod 998244353
#define m_p(a,b) make_pair(a, b)
#define mem(a,b) memset((a),(b),sizeof(a))
#define io ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
typedef long long ll;
typedef pair <int,int> pii;
#define MAX 300000 + 50
ll n, m, k, op;
int tot;
ll prim[MAX];
bool vis[1000050];
void init(){
for(int i = 2; i <= 1000000; ++i){
if(!vis[i])prim[++tot] = i;
for(int j = 1; j <= tot && prim[j] * i <= 1000000; ++j){
vis[prim[j] * i] = 1;
if(i % prim[j] == 0)break;
}
}
}
void work(){
init();
cin >> n;
int ans = 0;
for(int i = 1; i <= tot && (ll)pow(prim[i], 4) <= n; ++i){
int l = i + 1, r = tot;
while(l <= r){
int mid = (l + r) / 2;
if(prim[mid] * prim[mid] * prim[mid] > n / prim[i])r = mid - 1;
else l = mid + 1;
}
ans += l - i - 1;
}
cout << ans << endl;
}
int main(){
io;
work();
return 0;
}
J - Wrapping Chocolate
Title Description ：
n A chocolate ,m Boxes , The chocolates and boxes here are only long and wide , Ask if you can put n Chocolate is put in different boxes
Ideas ：
We will n+m Two chocolates and a box , According to the width, from big to small , The length is sorted from large to small , Then use one multiset To maintain that the unused width in the box is greater than or equal to all the lengths of the current chocolate , When it comes to chocolate , Just go to multiset Find his long in half , If found, delete , Output if you can't find it
No, If you encounter a box , Then the long race to multiset in
#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
#define inf 0x3f3f3f3f
#define mod 998244353
#define sd(n) scanf("%d",&n)
#define sdd(n,m) scanf("%d %d",&n,&m)
#define m_p(a,b) make_pair(a, b)
#define mem(a,b) memset((a),(b),sizeof(a))
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define io ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
typedef long long ll;
typedef pair <int,int> pii;
#define MAX 500000 + 50
int n, m, k, op;
struct ran{
int x, y;
bool op;
bool operator < (const ran &a)const{
if(x != a.x)return x > a.x;
else if(y != a.y)return y > a.y;
else return op > a.op;
}
}tr[MAX];
void work(){
cin >> n >> m;
for(int i = 1; i <= n; ++i){
cin >> tr[i].x;
}
for(int i = 1; i <= n; ++i){
cin >> tr[i].y;
}
for(int i = n + 1; i <= n + m; ++i){
cin >> tr[i].x;
}
for(int i = n + 1; i <= n + m; ++i){
cin >> tr[i].y;
tr[i].op = 1;
}
sort(tr + 1, tr + 1 + n + m);
multiset<int>se;
for(int i = 1; i <= n + m; ++i){
if(tr[i].op == 0){
auto x = se.lower_bound(tr[i].y);
if(x == se.end()){
cout << "No\n";
return;
}
se.erase(x);
}
else{
se.insert(tr[i].y);
}
}
cout << "Yes\n";
}
int main(){
io;
work();
return 0;
}
K - Endless Walk
Title Description ：
Here you are. n A little bit ,m side , Ask how many points exist that you can walk into a ring starting from this point
Ideas ：
Reverse map building and topology sorting , Output the number of points that are not pushed into the queue
#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
#define inf 0x3f3f3f3f
#define mod 998244353
#define sd(n) scanf("%d",&n)
#define sdd(n,m) scanf("%d %d",&n,&m)
#define m_p(a,b) make_pair(a, b)
#define mem(a,b) memset((a),(b),sizeof(a))
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define io ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
typedef long long ll;
typedef pair <int,int> pii;
#define MAX 300000 + 50
int n, m, k, op;
int in[MAX];
int x, y;
vector<int>G[MAX];
bool vis[MAX];
void work(){
cin >> n >> m;
for(int i = 1; i <= m; ++i){
cin >> x >> y;
G[y].push_back(x);
++in[x];
}
queue<int>q;
int ans = 0;
for(int i = 1; i <= n; ++i){
if(in[i] == 0){
q.push(i);
++ans;
}
}
while(!q.empty()){
int u = q.front();q.pop();
for(auto v : G[u]){
--in[v];
if(in[v] == 0){
q.push(v);
++ans;
}
}
}
cout << n - ans << endl;
}
int main(){
io;
work();
return 0;
}
L - Powerful array
Title Description ：
give n Number ,m Time to ask , Each inquiry gives
l, r, Suppose that there is k Species , The value isa[k], And each number appearsnum[k]Time , Then the value of the interval is ∑ i = 1 k n u m [ k ] ∗ n u m [ k ] ∗ a [ k ] \sum_{i=1}^{k}{num[k]*num[k] * a[k]} ∑i=1knum[k]∗num[k]∗a[k]Find the interval value of each query
Ideas ：
Mo team board
Just push the formula
#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
#define inf 0x3f3f3f3f
#define mod 998244353
#define m_p(a,b) make_pair(a, b)
#define mem(a,b) memset((a),(b),sizeof(a))
#define io ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
typedef long long ll;
typedef pair <int,int> pii;
#define MAX 300000 + 50
int n, m, k, op;
int block;
ll ar[MAX];
int get_block(int x){
return x / block;
}
struct ran{
int l, r, id;
bool operator < (const ran &a)const{
int x = get_block(l);
int y = get_block(a.l);
if(x != y)return x < y;
else return r < a.r;
}
}tr[MAX];
ll ans[MAX];
ll num[1000005];
ll cnt;
void add(int id){
cnt += ar[id] * (2 * num[ar[id]] + 1);
++num[ar[id]];
}
void del(int id){
cnt += ar[id] * (1 - 2 * num[ar[id]]);
--num[ar[id]];
}
void work(){
cin >> n >> m;
block = sqrt(n);
for(int i = 1; i <= n; ++i)cin >> ar[i];
for(int i = 1; i <= m; ++i){
cin >> tr[i].l >> tr[i].r;
tr[i].id = i;
}
sort(tr + 1, tr + 1 + m);
int l = 1, r = 0;
for(int i = 1; i <= m; ++i){
int s = tr[i].l, t = tr[i].r;
while(r < t)add(++r);
while(r > t)del(r--);
while(l < s)del(l++);
while(l > s)add(--l);
ans[tr[i].id] = cnt;
}
for(int i = 1; i <= m; ++i){
cout << ans[i] << endl;
}
}
int main(){
io;
work();
return 0;
}
summary
This competition has 4 individual abc The topic , I have done two of the sign in questions , Just write it and hand it in , The other two have met , But I didn't write , Just take advantage of this competition to make up
Write in the middle G The question is dp I wrote for a long time wa, I changed my posture and passed , In the afternoon, I did some research on the film , Find out according to your own dp By definition , You should write a sentence
dp[i][j] = dp[i - 1][j], But it wasn't written during the game , So I started to wa, Then I changed my posture and passedK I really didn't expect it to be so simple , I'm still wondering how dfs Can spend less time updating ,yj Say reverse mapping and run a topological sort , It makes sense at one point. I can only say
L I took a look at the question contest , I know it's a Mo team , But because I don't write data structure questions very much , I haven't written much about Mo team , I didn't write it at the beginning , later 9 I'll leave work directly when I have a question , Just forget this question , I just made up my findings sb topic
J The question about chocolate is really wonderful , Although I can tell at a glance that it's an order and then I'll find it in two , But I don't know how to write , Between me and djk Write C When yj Just write it down
C I want to run violently at the first sight bfs, But I didn't think about it at first , It turned out to be sb topic
A topic djk Say you can write a line segment tree , But when I looked at the maintenance section gcd, Go straight up st Watch , It is rare to meet one that can be used st Table questions , Don't waste
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