henan 5
B
签到 但是wa dzk见死不救捏 (下面是B题错误代码）
#include <bits/stdc++.h>
typedef long long ll;
void solve() {
int n;
std::cin >> n;
std::vector<ll> a(n);
for(int i = 0; i < n; i++) {
std::cin >> a[i];
}
ll L,R,res = 0,cur = 0, pre = 0,cnt = 0;
std::cin >> L >> R;
for(int i = 0; i < n; i++) {
if(a[i] < L) res += (L - a[i]);
if(a[i] > R) cnt += (a[i] - R);
if(L <= a[i] && a[i] <= R) {
cur += (a[i] - L);
pre += (R - a[i]);
}
}
ll ans = std::max(res,cnt);
if(ans == res) {
if(cur >= ans - cnt)
std::cout << ans << '\n';
else {
std::cout << -1 << '\n';
}
}
else {
if(pre >= ans - res)
std::cout << ans << '\n';
else {
std::cout << -1 << '\n';
}
}
return;
}
int main() {
std::ios::sync_with_stdio(false);
std::cin.tie(nullptr);
int _ = 1;
//std::cin >> _;
while(_ --) {
solve();
}
return 0;
}
#include <bits/stdc++.h>
typedef long long ll;
void solve() {
int n;
std::cin >> n;
std::vector<ll> a(n);
for(int i = 0; i < n; i++) {
std::cin >> a[i];
}
ll L,R,res = 0,sum = 0,cnt = 0;
std::cin >> L >> R;
for(int i = 0; i < n; i++) {
if(a[i] < L) res += (L - a[i]);
if(a[i] > R) cnt += (a[i] - R);
sum += a[i];
}
ll ans = std::max(res,cnt);
if(sum > n * R || sum < n * L) {
std::cout << -1 << '\n';
return;
}
std::cout << ans << '\n';
return;
}
int main() {
std::ios::sync_with_stdio(false);
std::cin.tie(nullptr);
int _ = 1;
//std::cin >> _;
while(_ --) {
solve();
}
return 0;
}
D
区间修改 单点查询 板子题 用的树状数组
#include <bits/stdc++.h>
typedef long long ll;
void solve() {
int n,now = 0;
std::cin >> n;
std::vector<ll> a(n+1),c(n+1);
auto lowbit = [&] (ll x) -> ll {
return x & (-x);
};
auto update = [&] (ll x, ll k) -> void {
for(ll i = x; i <= n; i += lowbit(i))
c[i] += k;
};
auto query = [&] (ll x) -> ll {
ll sum = 0;
for(ll i = x; i > 0; i -= lowbit(i))
sum += c[i];
return sum;
};
for(int i = 1; i <= n; i++) {
std::cin >> a[i];
update(i,a[i] - now);
now = a[i];
}
int m;
std::cin >> m;
for(ll i = 1,p,l,r,d,x; i <= m; i++) {
std::cin >> p;
if(p == 1) {
std::cin >> l >> r >> d;
update(l,d);
update(r+1,-d);
}
else {
std::cin >> x;
std::cout << query(x) << '\n';
}
}
return;
}
int main() {
std::ios::sync_with_stdio(false);
std::cin.tie(nullptr);
int _ = 1;
//std::cin >> _;
while(_ --) {
solve();
}
return 0;
}
E
线性筛 + 枚举
#include <bits/stdc++.h>
typedef long long ll;
const int N = 1e5 + 50;
bool not_prime[N];
ll plist[N], cnt;
void init()
{
for(int i = 2;i < N; i++)
{
if(not_prime[i] == 0) plist[++cnt] = i;
for(int j=1;j<=cnt;j++)
{
int x;
x = i * plist[j];
if(x > N) break;
not_prime[x] = 1;
if(i % plist[j] == 0) break;
}
}
}
void solve()
{
int n;
std::cin >> n;
bool ok = false;
std::vector<int> v;
//std::cout << cnt << '\n';
for(int i = 1; i <= cnt; i++) {
int t = plist[i];
for(int j = 1; j <= cnt; j ++) {
if(plist[j] < n - t && not_prime[n - t - plist[j]] == 0) {
std::cout << t << " " << plist[j] <<" "<< n - t - plist[j]<< "\n";
ok = true;
return;
}
}
}
if(!ok) std::cout << -1 << "\n";
return;
}
int main()
{
std::ios::sync_with_stdio(false);
std::cin.tie(nullptr);
init();
int _ = 1;
std::cin >> _ ;
while(_ --) {
solve();
}
return 0;
}
G
爬楼梯 递推+打表+取模
#include <bits/stdc++.h>
typedef long long ll;
const ll mod = 1e9 + 7;
const int N = 1e6 + 50;
ll f[N];
void solve() {
int n;
std::cin >> n;
f[1] = 1, f[2] = 2, f[3] = 4;
for(int i = 4; i <= n; i++) {
f[i] = (((f[i-1] + f[i-2]) % mod + f[i-3]) %mod + mod) % mod;
}
std::cout << f[n] << '\n';
return;
}
int main() {
std::ios::sync_with_stdio(false);
std::cin.tie(nullptr);
int _ = 1;
//std::cin >> _;
while(_ --) {
solve();
}
return 0;
}
H
查询区间最大值
\(RMQ\)板子
#include <bits/stdc++.h>
typedef long long ll;
const int N = 1e6 + 50;
typedef long long ll;
const int M = 2007;
int f[N][30];
void solve() {
int n;
std::cin >> n;
std::vector<int> a(n+1),Log(n+1);
for(int i = 1; i <= n; i++) std::cin >> a[i];
for(int i = 1; i <= n; i++) f[i][0] = a[i];
Log[1] = 0, Log[2] = 1;
for(int i = 3; i <= n; i++) Log[i] = Log[i/2] + 1;
for(int j = 1; j <= Log[n]; j ++) {
for(int i = 1; i + (1 << j) - 1 <= n; i++) {
f[i][j] = std::max(f[i][j-1],f[i + (1 <<(j-1))][j-1]);
}
}
int q;
std::cin >> q;
for(int i = 1; i <= q; i++) {
int l,r;
std::cin >> l >> r;
int p = Log[r - l + 1];
int ans = std::max(f[l][p],f[r - (1 << p) + 1][p]);
std::cout << ans << '\n';
}
return;
}
int main() {
std::ios::sync_with_stdio(false);
std::cin.tie(nullptr);
int _ = 1;
//std::cin >> _;
while(_ --) {
solve();
}
return 0;
}
I
三分 求极值
#include <bits/stdc++.h>
typedef long long ll;
const int N = 2e6 + 50;
void solve()
{
int n;
std::cin >> n;
std::vector<ll> a(n + 1), c(n + 1);
for(int i = 1; i <= n; i++) std::cin >> c[i];
auto check = [&](ll x) -> ll {
ll res = 0;
for(int i = 1; i <= n; i++) {
res += std::abs(c[i] - 1ll * x);
}
return res;
};
ll l = 0, r = 1e10;
ll ans = 1e14 + 7;
while(r - l > 1) {
ll mid = l + r >> 1;
ll lmid = mid - 1;
ll rmid = mid + 1;
if(check(lmid) < check(rmid)) r = mid;
else l = mid;
}
//std::cout << l << ' ' << r << '\n';
ans = std::min({ans,check(l),check(r)});
std::cout << ans + 1 << '\n';
return;
}
int main()
{
std::ios::sync_with_stdio(false);
std::cin.tie(nullptr);
int _ = 1;
//std::cin >> _ ;
while(_ --) {
solve();
}
return 0;
}
J
开方签到题
#include <bits/stdc++.h>
typedef long long ll;
void solve() {
ll n;
std::cin >> n;
std::cout << int(std::sqrt(n)) << '\n';
return;
}
int main() {
std::ios::sync_with_stdio(false);
std::cin.tie(nullptr);
int _ = 1;
std::cin >> _;
while(_ --) {
solve();
}
return 0;
}
K
s - t 的一条路径 要求路径的最大值最小
二分 + dijkstra判断联通
#include <bits/stdc++.h>
typedef long long ll;
const int N = 2e6 + 50;
const int INF = 0x3f3f3f3f;
struct edge{
int from;
int to;
int dis;
int next;
}e[N];
int head[N],cnt;
void add(int u,int v,int w)
{
++cnt;
e[cnt].from = u;
e[cnt].to = v;
e[cnt].dis = w;
e[cnt].next = head[u];
head[u] = cnt;
}
struct node{
int dis;
int pos;
bool operator <(const node &x)const
{
return x.dis < dis;
}
};
void solve()
{
int n,m;
std::cin >> n >> m;
std::vector<int> a(m),b(m),c(m);
for(int i = 0; i < m; ++ i) {
std::cin >> a[i] >> b[i] >> c[i];
}
int s,t;
std::cin >> s >> t;
auto check = [&](int x) -> bool {
for(int i = 0; i < N; i++) {
head[i] = 0;
e[i] = {0,0,0,0};
} cnt = 0;
for(int i = 0; i < m; ++ i) {
if(c[i] <= x) {
add(a[i], b[i], c[i]);
add(b[i], a[i], c[i]);
}
}
std::vector<int> f(n+1);
std::vector<bool> vis(n+1,false);
for(int i = 1; i <= n; ++ i) f[i] = INF;
f[s] = 0;
std::priority_queue<node> q;
q.push((node){0,s});
while(!q.empty()) {
node p = q.top();
q.pop();
int x = p.pos;
if(vis[x]) continue;
vis[x] = true;
for(int i = head[x]; i; i = e[i].next) {
int y = e[i].to;
if(f[y] > f[x] + e[i].dis) {
f[y] = f[x] + e[i].dis;
q.push((node){f[y],y});
}
}
}
return f[t] != INF;
};
int l = 0, r = 1e4, x = 0;
while(l <= r) {
int mid = (l + r) >> 1;
if(check(mid)) {
x = mid;
r = mid - 1;
}
else{
l = mid + 1;
}
}
std::cout << x << '\n';
return;
}
int main()
{
std::ios::sync_with_stdio(false);
std::cin.tie(nullptr);
int _ = 1;
//std::cin >> _ ;
while(_ --) {
solve();
}
return 0;
}
F
题目题意是 1 到各点最短距离 + 各点到1最短距离
但是 \(1\le n,m\le 10^6\) 单向边 n次dijkstra = \(n*n*log n\)
之前的 \(johnson\) 全源最短路 \(O(nmlogm)\) 也不行
\(Floyed\) 更是不行
so cant dzk ngmay
在dzk 的亲切指导下，我动力只要建反向边 再跑一遍就是各点到 1 的距离
#include <bits/stdc++.h>
typedef long long ll;
const int N = 2e6 + 50;
const int INF = 0x3f3f3f3f;
struct node{
int dis;
int pos;
bool operator <(const node &x) const {
return x.dis < dis;
}
};
struct edge{
int to;
int dis;
int next;
}e[N];
int head[N],cnt;
void add(int u,int v,int w){
e[++cnt] = (edge){v,w,head[u]};
head[u] = cnt;
}
void solve()
{
int n,m;
std::cin >> n >> m;
std::vector<int> a(m),b(m),c(m);
for(int i = 0;i < m;++i) {
std::cin >> a[i] >> b[i] >> c[i];
}
ll ans = 0;
auto Dijkstra = [&](int s) -> void{
std::vector<int> f(n+1);
std::vector<bool> vis(n+1,false);
for(int i = 1; i <= n; ++ i) f[i] = INF;
f[s] = 0;
std::priority_queue<node> q;
q.push((node){0,s});
while(!q.empty()) {
node p = q.top();
q.pop();
int x = p.pos;
if(vis[x]) continue;
vis[x] = true;
for(int i = head[x]; i; i = e[i].next) {
int y = e[i].to;
if(f[y] > f[x] + e[i].dis) {
f[y] = f[x] + e[i].dis;
q.push((node){f[y],y});
}
}
}
for(int i = 2; i <= n; i++) ans += 1ll * f[i];
};
for(int i = 0;i < m;++i) {
add(a[i],b[i],c[i]);
}
Dijkstra(1);
for(int i = 0; i <= cnt; i ++) {
e[i] = {0,0,0};
} std::memset(head,0,sizeof(head)); cnt = 0;
for(int i = 0;i < m;++i) {
add(b[i],a[i],c[i]);
}
Dijkstra(1);
std::cout << ans << '\n';
return;
}
int main()
{
std::ios::sync_with_stdio(false);
std::cin.tie(nullptr);
int _ = 1;
//std::cin >> _ ;
while(_ --) {
solve();
}
return 0;
}
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