A.
Example
Input
9
1 3 3 6 7 6 8 8 9
Output
4
1 #include <iostream> 2 3 using namespace std; 4 5 int main () { 6 int n; 7 cin >> n; 8 int a[10010]; 9 for(int i = 0;i < n;i++){ 10 cin >> a[i]; 11 } 12 int day = 0; 13 int t = 0; 14 for(int i = 1;i <= n;i++){ 15 t = max(a[i - 1],t); 16 if(i == t) 17 day++; 18 } 19 cout << day << endl; 20 return 0; 21 }
B
Example
Input
3
2
<>
3
><<
1
>
Output
1
0
0
1 // 本题问的是先删除后再按照上述判断 进而看是否能构成 2 #include <bits/stdc++.h> 3 4 using namespace std; 5 6 int main(){ 7 ios_base::sync_with_stdio(0); 8 cin.tie(0); 9 cout.tie(0); 10 int t; 11 cin >> t; 12 while (t--) { 13 int n; 14 cin >> n; 15 string s; 16 cin >> s; 17 int l = 0; 18 int r = n - 1; 19 while(l < n && s[l] == '<') 20 l++; 21 cout << "l :" << l << endl; 22 while(r >= 0 && s[r] == '>') 23 r--; 24 cout << "r " << r << endl; 25 cout << min(l, n - r - 1) << endl; 26 // 从左边开始‘<’的连续个数与从右边开始‘>’的连续个数 两者的最小值即为答案 27 // 因为除去这个边界之后的下一个数一定与它相反 通过它就可以实现转化 28 } 29 }
C
Examples
Input
4 3
4 7
15 1
3 6
6 8
Output
78
Input
5 3
12 31
112 4
100 100
13 55
55 50
Output
10000
#include<bits/stdc++.h>
using namespace std;
const int N = 3e5+1;
long long n, k, x, y, s, A;
pair<int,int>p[N];
priority_queue<int>q;
// 优先队列就是大顶堆，队头元素最大
int main(){
cin >> n >> k;
for (int i = 1; i <= n; i++) { // 从第一个开始进行存入
cin >> x >> y;
p[i] = make_pair(y, x);
}
sort(p + 1, p + n + 1); // pair 默认对first升序，当first相同时对second升序（就是对beauty values进行排序）
for(int i = n; i >= 1; i--) {
s += p[i].second; // 将 lengths 从大到小相加
q.push(-p[i].second); // 将length的相反数放入队列中
if (q.size() > k) { // 当压入的数目超过最大个数时
s += q.top(); // s加上最上面压入的元素（正值最小的）
q.pop(); // 弹出最上面的
}
A = max(A, p[i].first * s); // p[i].first一定是该状态下的最大值
}
cout << A << endl;
return 0;
}
D
// 一个多变形最多能被分成 n - 2个不重合的三角形
#include <bits/stdc++.h>
using namespace std;
int main () {
int n;
cin >> n;
int sum = 0;
for (int i = 2;i < n; i++) {
sum += i * (i + 1);
}
cout << sum << endl;
}
待补！！！
E
#include <bits/stdc++.h>
using namespace std;
#define MOD 998244353
int n, k, x[2], at;
long long dp[2][2];
int ta;
int main() {
scanf("%d%d", &n, &k);
for (int i = 0; i < n; i++) {
at = i & 1;
scanf("%d", &ta);
if (ta == -1) {
if (x[at] == 0) {
x[at] = 1;
dp[at][0] = dp[at][1] = 1;
} else {
long long tmp = dp[at][0];
dp[at][0] = (dp[at][0] * (k - 2) + dp[at][1]) % MOD;
dp[at][1] = tmp * (k - 1) % MOD;
}
} else {
if (x[at] == 0)
dp[at][1] = 1;
else if (ta == x[at])
dp[at][1] = dp[at][0] * (k - 1) % MOD;
else
dp[at][1] = (dp[at][0] * (k - 2) + dp[at][1]) % MOD;
dp[at][0] = 0;
x[at] = ta;
}
}
for (int i = 0; i < 2; i++)
dp[i][0] = (dp[i][0] * (k - 1) + dp[i][1]) % MOD;
printf("%lld\n", dp[0][0] * dp[1][0] % MOD);
}
F
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define f(i, x, n) for (int i = x; i < (int)(n); ++i)
int const N = 300000;
int q;
vector<pair<int, int> > s[N << 2];
map<pair<int, int>, int> edtm;
vector<pair<int, int> > p;
ll an[N + 1];
void adsg(pair<int, int> e, int x, int y, int l = 1, int r = q, int id = 1){
if (l > y || r < x)return;
if (l >= x && r <= y) { s[id].push_back(e); return; }
int m = l + r >> 1, a = id << 1, b = a | 1;
adsg(e, x, y, l, m, a);
adsg(e, x, y, m + 1, r, b);
}
int pa[N << 1 | 1];
pair<int, int> sz[N << 1 | 1];
stack<pair<int, int> > st;
ll can;
int P(int p) { while (pa[p])p = pa[p]; return p; }
void adan(pair<int, int> p) { can += (ll)p.first * p.second; }
void sban(pair<int, int> p) { can -= (ll)p.first * p.second; }
void fn(int l = 1, int r = q, int id = 1){
int undc = 0;
for (auto &t: s[id]){
int a = P(t.first), b = P(t.second);
if (a == b)continue;
if (sz[a].first + sz[a].second < sz[b].first + sz[b].second)swap(a, b);
pa[b] = a;
sban(sz[a]);
sban(sz[b]);
sz[a].first += sz[b].first;
sz[a].second += sz[b].second;
adan(sz[a]);
st.push(make_pair(a, b));
++undc;
}
if (l == r)an[l] += can;
else {
int m = l + r >> 1, a = id << 1, b = a | 1;
fn(l, m, a);
fn(m + 1, r, b);
}
while (undc--){
int a = st.top().first, b = st.top().second;
st.pop();
sban(sz[a]);
sz[a].first -= sz[b].first;
sz[a].second -= sz[b].second;
adan(sz[a]);
adan(sz[b]);
pa[b] = 0;
}
}
G
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define f(i, x, n) for (int i = x; i < (int)(n); ++i)
ll const inf = 1e18;
int const N = 300000, B = 19;
ll wsd[N + 1], w[N + 1][B][2][2];
int pa[N + 1][B], dp[N + 1];
struct E{
int u;
ll w[2];
E(int a, ll b, ll c):u(a) { w[0] = b, w[1] = c; }
};
vector<E> tr[N + 1];
struct D{
ll w[2];
D(int v) { w[0] = w[1] = 0; w[v & 1] = inf; }
void operator +=(ll c[][2]){
ll nw0 = min(w[0] + c[0][0], w[1] + c[1][0]);
w[1] = min(w[0] + c[0][1], w[1] + c[1][1]);
w[0] = nw0;
}
};
void J(ll a[][2], ll b[][2], ll c[][2]){
f(i, 0, 2)f(j, 0, 2)c[i][j] = inf;
f(i, 0, 2)f(j, 0, 2)f(k, 0, 2)c[i][j] = min(c[i][j], a[i][k] + b[k][j]);
}
void pl1(int v = 1, int p = 0){
for (E &t: tr[v])if (t.u != p)pl1(t.u, v);
for (E &t: tr[v])if (t.u != p)wsd[v] = min(wsd[v], wsd[t.u] + t.w[0] + t.w[1]);
}
void pl2(int v = 1, int p = 0) { for (E &t: tr[v])if (t.u != p)wsd[t.u] = min(wsd[t.u], wsd[v] + t.w[0] + t.w[1]), pl2(t.u, v); }
void pl3(int v = 1, int p = 0){
pa[v][0] = p;
dp[v] = dp[p] + 1;
int k = 1, z;
while (k < B && (z = pa[v][k - 1])){
pa[v][k] = pa[z][k - 1];
J(w[v][k - 1], w[z][k - 1], w[v][k]);
++k;
}
for (E &t: tr[v]){
if (t.u == p)continue;
w[t.u][0][0][0] = min(t.w[0], t.w[1] + wsd[v] + wsd[t.u]);
w[t.u][0][0][1] = min(t.w[0] + wsd[v], wsd[t.u] + t.w[1]);
w[t.u][0][1][0] = min(t.w[1] + wsd[v], wsd[t.u] + t.w[0]);
w[t.u][0][1][1] = min(t.w[1], t.w[0] + wsd[v] + wsd[t.u]);
pl3(t.u, v);
}
}
ll wan[2][2];
ll go(int a, int b){
D ca(a), cb(b);
a = a + 1 >> 1, b = b + 1 >> 1;
if (dp[a] < dp[b])swap(a, b), swap(cb, ca);
int d = dp[a] - dp[b], j = 0;
while (d){
if (d & 1)ca += w[a][j], a = pa[a][j];
++j;
d >>= 1;
}
if (a != b){
for (int i = B - 1; i >= 0; --i)if (pa[a][i] != pa[b][i])ca += w[a][i], cb += w[b][i], a = pa[a][i], b = pa[b][i];
ca += w[a][0], cb += w[b][0];
a = pa[a][0];
}
ll an = inf;
f(i, 0, 2)f(j, 0, 2)an = min(an, ca.w[i] + cb.w[j] + (i != j ? wsd[a] : 0));
return an;
}
int main(){
int n;
scanf("%d", &n);
f(i, 1, n + 1)scanf("%lld", wsd + i);
f(i, 1, n){
int a, b;
ll w1, w2;
scanf("%d%d%lld%lld", &a, &b, &w1, &w2);
tr[a].push_back(E(b, w1, w2));
tr[b].push_back(E(a, w1, w2));
}
pl1();
pl2();
pl3();
int q;
scanf("%d", &q);
while (q--){
int a, b;
scanf("%d%d", &a, &b);
printf("%lld\n", go(a, b));
}
}