A - 欢迎来到辽宁省赛
#include <bits/stdc++.h>
using namespace std;
int main(){
cout << "27\n";
return 0;
}
B - 胜率
首先如果参加了 10000 局比赛，则一定会有当前胜率的情况出现。
所以我们枚举一下分母就好了
#include <bits/stdc++.h>
using namespace std;
int main(){
double A;
cin >> A, A /= 100.0;
for(int x , y = 1; y <= 100000; y ++) {
x = A * y;
for(int i = max(x - 10 , 0) , v; i <= min(x + 10 , y); i ++){
v = (double(i) / double(y) * 1e5 + 5) / 10;
//v /= 1000;
if(abs(A * 100 - double(v) / 100) == 0) {
cout << y << "\n";
return 0;
}
}
}
return 0;
}
C - 连环爆炸
先删掉所有无法被炸死的怪，然后贪心的删除爆炸伤害高的。
#include <bits/stdc++.h>
using namespace std;
using pii = pair<int,int>;
bool cmp(const pii &x , const pii &y) {
if(x.second != y.second) return x.second < y.second;
return x.first < y.first;
}
int main() {
int n, res = 0, sum = 0;
cin >> n;
vector<pii> f(n);
for(auto &[a,b] : f) cin >> a >> b , sum += b;
{ // 删掉无法被炸死的
vector<pii> g;
queue<int> q;
q.push(0);
for(auto it : f){
if( it.first > sum - it.second) res ++, q.back() += it.second;
else g.push_back(it);
}
f = move(g);
while(not q.empty()) {
int x = q.front();
q.pop();
vector<pii> g;
for(auto &it : f){
it.first -= x;
if(it.first <= 0) q.push(it.second);
else g.push_back(it);
}
f = move(g);
}
}
while(not f.empty()) {
res ++;
queue<int> q;
sort(f.begin(), f.end(), cmp);
q.push(f.back().second), f.pop_back();
while(not q.empty()) {
int x = q.front();
q.pop();
vector<pii> g;
for(auto &it : f){
it.first -= x;
if(it.first <= 0) q.push(it.second);
else g.push_back(it);
}
f = move(g);
}
}
cout << res << "\n";
return 0;
}
F - 隔板与水槽
枚举中间点，贪心
#include <bits/stdc++.h>
using namespace std;
using i32 = int32_t;
using i64 = long long;
#define int i64
using vi = vector<int>;
i32 main(){
int n;
cin >> n;
vi h(n + 1);
for(int i = 1 ; i <= n; i ++)
cin >> h[i];
int res = 0;
for(int mid = 2 , l , r ; mid < n ; mid ++) {
l = r = 0;
for(int i = 1; i < mid ; i ++)
l = max(l, (mid - i) * min(h[i], h[mid]));
for(int i = mid + 1; i <= n; i ++)
r = max(r, (i - mid) * min(h[i], h[mid]));
res = max(res, l + r);
}
cout << res << "\n";
return 0;
}
H - 取石子
打表找规律，发现结果之和奇偶有关，因为 Alice 和 Bob 都只能取奇数个。
#include <bits/stdc++.h>
using namespace std;
void solve() {
int a,b,c;cin >> a >> b >> c;
a %= 2,b %= 2,c %= 2;
int ans = (a << 2) + (b << 1) + (c << 0);
if (ans % 2) cout << "Alice\n";
else
cout << "Bob\n";
}
int main(){
ios::sync_with_stdio(false);
cin.tie(nullptr);
int TC;
for(cin >> TC; TC; TC --)
solve();
return 0;
}
M - 让二追三
每个位置出现复合条件字符串的是独立的。
#include <bits/stdc++.h>
using namespace std;
using i32 = int32_t;
#define int long long
const int mod = 1e9+7;
int power(int x, int y) {
int ans = 1;
while(y) {
if(y & 1) ans = ans * x % mod;
x = x * x % mod, y /= 2;
}
return ans;
}
int inv(int x) {
return power(x, mod - 2);
}
void solve(){
int n, a, b;
cin >> a >> b >> n;
if(n < 5) {
cout << "0\n";
return;
}
int p = a * inv(b) % mod;
p = (1 - p + mod ) % mod * (1 - p + mod) % mod * p % mod * p %mod * p % mod;
cout << (n - 4) * p % mod << "\n";
}
i32 main(){
ios::sync_with_stdio(false), cin.tie(nullptr);
int TC;
for(cin >> TC; TC; TC -- )
solve();
return 0;
}
L - 区间与绝对值
队友写的莫队
#include <bits/stdc++.h>
using namespace std;
#define ll long long
const int N = 1e5 + 7;
template<class T>
struct Fenwick{
vector<T> c;
int n;
Fenwick(int _n){
n = _n;
c.resize(n + 1);
}
T sum(int x){
T res = 0;
for (; x ; x -= x & (-x)){
res += c[x];
}
return res;
}
void modify(int x,T d){
for (; x <= n ;x += x & (-x)){
c[x] += d;
}
}
};
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n,m;cin >> n >> m;
vector<int> a(n + 1);
vector<ll> res(m + 1);
Fenwick<ll> pre(N);
Fenwick<ll> siz(N);
for (int i = 1;i <= n;i++){
cin >> a[i];
}
vector<array<int,3>> qury;
for (int i = 1;i <= m;i++){
int l,r;cin >> l >> r;
qury.push_back({l,r,i});
}
int B = 500;
sort(qury.begin(),qury.end(),[&](array<int,3> a,array<int,3> b){
int c = a[0] / B;
if (a[0] / B != b[0] / B) return a[0] / B < b[0] / B;
return c % 2 ? a[1] < b[1] : a[1] > b[1];
});
int l = 1,r = 0;
ll ans = 0;
auto add = [&](int x)->void{
ll p = a[x] * (siz.sum(a[x])) - pre.sum(a[x]);
ll s = (pre.sum(1e5) - pre.sum(a[x])) - (siz.sum(1e5) - siz.sum(a[x])) * a[x];
ans += p + s;
pre.modify(a[x],a[x]);
siz.modify(a[x],1);
//cout << ans << endl;
};
auto del = [&](int x)->void{
ll p = a[x] * (siz.sum(a[x])) - pre.sum(a[x]);
ll s = (pre.sum(1e5) - pre.sum(a[x])) - (siz.sum(1e5) - siz.sum(a[x])) * a[x];
ans -= p + s;
siz.modify(a[x],-1);
pre.modify(a[x],-a[x]);
};
for (int i = 0;i < m;i++){
while (r < qury[i][1]) r++,add(r);
while (l > qury[i][0]) l--,add(l);
while (r > qury[i][1]) del(r),r--;
while (l < qury[i][0]) del(l),l++;
res[qury[i][2]] = ans;
}
for (int i = 1;i <= m;i++){
cout << 2 * res[i] << "\n";
}
return 0;
}